Binomial Theorem and its Simple Applications jee mains questions | Binomial Theorem and its Simple Applications Previous Year Question

Q 1 :

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})}^{n}$ is $\sqrt{6} : 1$ , then the third term from the beginning is [2023]

$30 \sqrt{2}$
$60 \sqrt{3}$
$60 \sqrt{2}$
$30 \sqrt{3}$

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(2)

Q 2 :

If the coefficients of $x^{7}$ in ${(a x^{2} + \frac{1}{2 b x})}^{11}$ and $x^{- 7}$ in ${(a x - \frac{1}{3 b x^{2}})}^{11}$ are equal, then [2023]

64ab = 243
729ab = 32
32ab = 729
243ab = 64

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(2)

We have, $r^{t h}$ term in ${(a x^{2} + \frac{1}{2 b x})}^{11}$ is,

$T_{r + 1} = {}^{11}{C_{r}} {(a x^{2})}^{11 - r} {(\frac{1}{2 b x})}^{r} = {}^{11}{C_{r}} \frac{a^{11 - r}}{{(2 b)}^{r}} \cdot x^{22 - 3 r}$

For coefficient of $x^{7}$ , we have $22 - 3 r = 7 \Rightarrow r = 5$

$∴$ $T_{6} = {}^{11}{C_{5}} \frac{a^{6}}{2^{5} b^{5}} x^{7} ...(i)$

Similarly, $r^{t h}$ term in the second expansion is, $T_{r + 1} = {}^{11}{C_{r}} {(a x)}^{11 - r} {(\frac{- 1}{3 b x^{2}})}^{r} = {}^{11}{C_{r}} \frac{a^{11 - r}}{{(- 3 b)}^{r}} \cdot x^{11 - 3 r}$

For coefficient of $x^{- 7}$ , we have $11 - 3 r = - 7 \Rightarrow r = 6$

$∴$ $T_{7} = {}^{11}{C_{6}} \frac{a^{5}}{3^{6} \cdot b^{6}} x^{- 7} ...(ii)$

Now, ${}^{11}{C_{5}} \frac{a^{6}}{2^{5} b^{5}} = {}^{11}{C_{6}} \frac{a^{5}}{3^{6} b^{6}}$ $\Rightarrow 3^{6} a b = 2^{5} \Rightarrow 729 a b = 32$

Q 3 :

If the coefficients of three consecutive terms in the expansion of ${(1 + x)}^{n}$ are in the ratio 1:5:20, then the coefficient of the fourth term is [2023]

1827
5481
2436
3654

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(4)

Given, coefficients of three consecutive terms are in the ratio 1 : 5 : 20

$i.e., {}^{n}{C_{r - 1}} : {}^{n}{C_{r}} : {}^{n}{C_{r + 1}} = 1 : 5 : 20$

Now, $\frac{{}^{n}{C_{r - 1}}}{{}^{n}{C_{r}}} = \frac{1}{5} \Rightarrow n = 6 r - 1 ...(i)$

Also, $\frac{{}^{n}{C_{r}}}{{}^{n}{C_{r + 1}}} = \frac{5}{20} \Rightarrow n = 5 r + 4 ...(ii)$

From (i) and (ii), we get

$6 r - 1 = 5 r + 4 \Rightarrow r = 5 \Rightarrow n = 25 + 4 = 29$

$∴$ $The coefficient of 4th term = {}^{29}{C_{3}} = 3654$

Q 4 :

The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of ${(2 x^{2} + \frac{1}{2 x})}^{11}$ is equal to [2023]

$10^{3} - 10$
$12^{3} - 12$
$13^{3} - 13$
$11^{3} - 11$

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(2)

We have, $T_{r + 1} = {}^{11}{C_{r}} {(2 x^{2})}^{11 - r} {(\frac{1}{2 x})}^{r} = {}^{11}{C_{r}} 2^{11 - 2 r} \cdot x^{22 - 3 r}$

For $x^{10}$ , $22 - 3 r = 10 \Rightarrow r = 4$

$∴$ $Coefficient of x^{10} = {}^{11}{C_{4}}$ $2^{11 - 8} = 330 \times 2^{3} = 2640$

Now, for $x^{7}$ , $22 - 3 r = 7 \Rightarrow r = 5$

$∴$ $Coefficient of x^{7} = {}^{11}{C_{5}} \cdot 2^{11 - 10} = 462 \times 2 = 924$

So, absolute difference of the coefficients of $x^{10}$ and $x^{7}$ is

$2640 - 924 = 1716 = 12^{3} - 12$

Q 5 :

If the coefficient of $x^{7}$ in ${(a x - \frac{1}{b x^{2}})}^{13}$ and the coefficient of $x^{- 5}$ in ${(a x + \frac{1}{b x^{2}})}^{13}$ are equal, then $a^{4} b^{4}$ is equal to [2023]

11
44
22
33

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(3)

In the first expansion,
$T_{r + 1} = {}^{13}{C_{r}} {(a x)}^{13 - r} {(\frac{- 1}{b x^{2}})}^{r} = {}^{13}{C_{r}} a^{13 - r} {(\frac{- 1}{b})}^{r} x^{13 - 3 r}$

Now, $13 - 3 r = 7 \Rightarrow r = 2$

$∴$ $Coefficient of x^{7} = {}^{13}{C_{2}} \frac{a^{11}}{b^{2}}$

Also, in the second expansion,
$T_{r + 1} = {}^{13}{C_{r}} {(a x)}^{13 - r} {(\frac{1}{b x^{2}})}^{r} = {}^{13}{C_{r}} \frac{a^{13 - r}}{b^{r}} x^{13 - 3 r}$

As, $13 - 3 r = - 5 \Rightarrow r = 6$

$∴$ $Coefficient of x^{- 5} = {}^{13}{C_{6}} \frac{a^{7}}{b^{6}}$

Now, ${}^{13}{C_{2}} \frac{a^{11}}{b^{2}} = {}^{13}{C_{6}} \frac{a^{7}}{b^{6}}$ $\Rightarrow a^{4} b^{4} = \frac{{}^{13}{C_{6}}}{{}^{13}{C_{2}}} = 22$

Q 6 :

If the coefficients of $x$ and $x^{2}$ in ${(1 + x)}^{p} {(1 - x)}^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to [2023]

66
60
63
69

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(3)

${(1 + x)}^{p} {(1 - x)}^{q} = [1 + p x + \frac{p (p - 1)}{2} x^{2} + \dots] [1 - q x + \frac{q (q - 1)}{2} x^{2} + \dots]$

$∴$ $Coefficient of x = p - q = 4 ...(i)$

Coefficient of $x^{2}$ = $\frac{q (q - 1)}{2} - p q + \frac{p (p - 1)}{2} = - 5$

$\Rightarrow {(p - q)}^{2} - (p + q) = - 10 \Rightarrow p + q = 26 ...(ii)$

So, from (i) and (ii),

$p = 15, q = 11$ $∴$ $2 p + 3 q = 30 + 33 = 63$

Q 7 :

The sum of the coefficients of three consecutive terms in the binomial expansion of ${(1 + x)}^{n + 2}$ , which are in the ratio 1:3:5, is equal to [2023]

41
92
25
63

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(4)

Given ${}^{n + 2}{C_{r - 1}} : {}^{n + 2}{C_{r}} : {}^{n + 2}{C_{r + 1}} = 1 : 3 : 5$

$∴$ $\frac{{}^{n + 2}{C_{r - 1}}}{{}^{n + 2}{C_{r}}} = \frac{1}{3} \Rightarrow n = 4 r - 3 ...(i)$

Also, $\frac{{}^{n + 2}{C_{r}}}{{}^{n + 2}{C_{r + 1}}} = \frac{3}{5} \Rightarrow 8 r - 1 = 3 n ...(ii)$

From (i) and (ii), we get

$4 r - 3 = \frac{8 r - 1}{3} \Rightarrow 4 r = 8 \Rightarrow r = 2 and n = 5$

$Required sum = {}^{7}{C_{1}} + {}^{7}{C_{2}} + {}^{7}{C_{3}} = 7 + 21 + 35 = 63$

Q 8 :

The coefficient of $x^{5}$ in the expansion of ${(2 x^{3} - \frac{1}{3 x^{2}})}^{5}$ is [2023]

$\frac{26}{3}$
9
8
$\frac{80}{9}$

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(4)

${(2 x^{3} - \frac{1}{3 x^{2}})}^{5}$

$T_{r + 1} = {}^{5}C_{r} {(2 x^{3})}^{5 - r} {(\frac{- 1}{3 x^{2}})}^{r} = {}^{5}C_{r} \cdot 2^{5 - r} \cdot {(\frac{- 1}{3})}^{r} \cdot x^{15 - 3 r - 2 r}$

$As, 15 - 5 r = 5 \Rightarrow r = 2$

$Coefficient of x^{5} = {}^{5}C_{2} 2^{3} \times {(\frac{- 1}{3})}^{2} = \frac{80}{9}$

Q 9 :

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of ${(1 + x)}^{10}$ , then $\sum_{r = 1}^{10} r^{3} {(\frac{a_{r}}{a_{r - 1}})}^{2}$ is equal to [2023]

3025
1210
5445
4895

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(2)

Q 10 :

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of ${(1 + x)}^{99}$ . Let $a$ be the middle term in the expansion of ${(2 + \frac{1}{\sqrt{2}})}^{200}$ .

If $\frac{{}^{200}{C_{99} K}}{a} = \frac{2^{l} m}{n}$ , where $m$ and $n$ are odd numbers, then the ordered pair $(l, n)$ is equal to [2023]

(50, 101)
(51, 101)
(51, 99)
(50, 51)

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(1)

$Sum of the coefficients of odd powers of x in the expansion {(1 + x)}^{99} be K .$

$If a be the middle term in expansion of {(2 + \frac{1}{\sqrt{2}})}^{200} .$

$In the expansion of {(1 + x)}^{99} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{99} x^{99}$

$If a = middle term of$ ${(2 + \frac{1}{\sqrt{2}})}^{200}$

$= T_{(\frac{200}{2} + 1)} = {}^{200}C_{100} {(2)}^{100} {(\frac{1}{\sqrt{2}})}^{100} \Rightarrow T_{101} = {}^{200}C_{100} \cdot 2^{50}$

$K = 2^{98}; T_{101} = {}^{200}C_{100} \cdot 2^{50} = a$

$So, \frac{{}^{200}C_{99} \times 2^{98}}{{}^{200}C_{100} \times 2^{50}} = \frac{100}{101} \times 2^{48} \Rightarrow \frac{25}{101} \times 2^{50} = \frac{m}{n} 2^{l}$

$∴$ $m, n are odd, so (l, n) becomes (50, 101)$

Q 11 :

The coefficient of $x^{301}$ in ${(1 + x)}^{500} + x {(1 + x)}^{499} + x^{2} {(1 + x)}^{498} + . . . + x^{500}$ is [2023]

${}^{500}{C_{301}}$
${}^{501}{C_{200}}$
${}^{500}{C_{300}}$
${}^{501}{C_{302}}$

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(2)

$The coefficient of x^{301} in$

${(1 + x)}^{500} + x {(1 + x)}^{499} + x^{2} {(1 + x)}^{498} + \dots + x^{500}$

$= {}^{500}C_{301} + {}^{499}C_{300} + {}^{498}C_{299} + \dots + {}^{199}C_{0}$

$= {}^{500}C_{199} + {}^{499}C_{199} + {}^{498}C_{199} + \dots + {}^{199}C_{199} = {}^{501}C_{200}$

Q 12 :

If the coefficient of $x^{15}$ in the expansion of ${(a x^{3} + \frac{1}{b x^{1 / 3}})}^{15}$ is equal to the coefficient of $x^{- 15}$ in the expansion of ${(a x^{1 / 3} - \frac{1}{b x^{3}})}^{15}$ , where $a$ and $b$ are positive real numbers, then for each such ordered pair (a, b) [2023]

a = 3b
a = b
ab = 1
ab = 3

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(3)

The coefficient of $x^{15}$ in the expansion of

${(a x^{3} + \frac{1}{b x^{\frac{1}{3}}})}^{15}$ is $T_{r + 1} = {}^{15}{C_{r}} {(a x^{3})}^{15 - r} \times {(b^{- 1} x^{- \frac{1}{3}})}^{r}$

$= {}^{15}{C_{r}} a^{15 - r} \cdot x^{45 - 3 r} \cdot b^{- r} \cdot x^{- \frac{r}{3}} = {}^{15}{C_{r}} a^{15 - r} \cdot b^{- r} x^{45 - 3 r - \frac{r}{3}}$

We have to find the coefficient of $x^{15}$ , so $x^{45 - 3 r - \frac{r}{3}} = x^{15}$

$\Rightarrow 45 - 3 r - \frac{r}{3} = 15 \Rightarrow \frac{10 r}{3} = 30 \Rightarrow r = 9$

So, $T_{9 + 1} = {}^{15}{C_{9}} a^{15 - 9} \cdot b^{- 9} \cdot x^{15};$ $T_{10} = {}^{15}{C_{9}} a^{6} \cdot b^{- 9} \cdot x^{15}$

$∴$ $The coefficient of x^{15} = {}^{15}{C_{9}} a^{6} \cdot b^{- 9}$

Now, the coefficient of $x^{- 15}$ in the expansion of

${({a x}^{\frac{1}{3}} - \frac{1}{b x^{3}})}^{15}$ is, $T_{r + 1} = {{}^{15}C}_{r} {(a x^{1 / 3})}^{15 - r} \cdot {(\frac{- 1}{b x^{3}})}^{r}$

$= {{}^{15}C}_{r} a^{15 - r} x^{\frac{15 - r}{3}} \cdot b^{- r} \cdot x^{- 3 r} {(- 1)}^{r}$

$= {}^{15}{C_{r}} a^{15 - r} \cdot b^{- r} \cdot x^{\frac{15 - r}{3} - 3 r} {(- 1)}^{r}$

We need to find the coefficient of $x^{- 15}$ , so

$x^{\frac{15 - r}{3} - 3 r} = x^{- 15} \Rightarrow \frac{15 - r}{3} - 3 r = - 15 \Rightarrow r = 6$

$∴$ $The coefficient of x^{- 15} = {}^{15}{C_{6}} a^{9} b^{- 6}$

Now, according to the question, ${}^{15}{C_{9}} a^{6} b^{- 9} = {}^{15}{C_{6}} a^{9} b^{- 6}$

$\Rightarrow a^{3} b^{3} = 1 ∴ a b = 1$

Q 13 :

The coefficient of $x^{18}$ in the expansion of ${(x^{4} - \frac{1}{x^{3}})}^{15}$ is __________ . [2023]

(5005)

Given, ${(x^{4} - \frac{1}{x^{3}})}^{15}$

$T_{r + 1} = {}^{15}{C_{r}} {(x^{4})}^{15 - r} {(\frac{- 1}{x^{3}})}^{r}$

$= {}^{15}{C_{r}} {(x)}^{60 - 4 r} \cdot {(- 1)}^{r} x^{- 3 r} = {}^{15}{C_{r}} {(x)}^{60 - 7 r} \cdot {(- 1)}^{r}$

For the coefficient of $x^{18}$

$60 - 7 r = 18 \Rightarrow r = \frac{42}{7} = 6$

Hence, coefficient of $x^{18}$ is ${}^{15}{C_{6}} {(- 1)}^{6} = 5005$

Q 14 :

If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]

(1275)

Let the $r^{t h}$ term be the constant term of the given expansion.

$∴$ $T_{r + 1} = {}^{7}{C_{r}} {(3 x^{2})}^{7 - r} \cdot {(\frac{- 1}{2 x^{5}})}^{r} = {}^{7}{C_{r}} 3^{7 - r} \cdot {(\frac{- 1}{2})}^{r} \cdot x^{14 - 2 r - 5 r}$

For constant term, put $14 - 2 r - 5 r = 0 \Rightarrow 7 r = 14 \Rightarrow r = 2$

$∴$ $T_{2 + 1} = {}^{7}{C_{2}} 3^{7 - 2} {(\frac{- 1}{2})}^{2} = 21 \times 3^{5} \times \frac{1}{4}$

$\Rightarrow α = 1275.75 \Rightarrow [α] = 1275$

Q 15 :

The number of integral terms in the expansion of ${(3^{1 / 2} + 5^{1 / 4})}^{680}$ is equal to ___________ . [2023]

(171)

The expansion of the general term is

$= {}^{680}{C_{r}} {(3^{1 / 2})}^{680 - r} {(5^{1 / 4})}^{r} = {}^{680}{C_{r}} 3^{\frac{680 - r}{2}} 5^{\frac{r}{4}}$

Possible values of $r$ , where $\frac{r}{4}$ is an integer

$r = 0, 4, 8, 12, \dots, 680$

All these values of $r$ are accepted as well.

$Hence, number of integral terms = 171$

Q 16 :

Let $α$ be the constant term in the binomial expansion of ${(\sqrt{x} - \frac{6}{x^{\frac{3}{2}}})}^{n}, n \leq 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{- n}$ is $λ α$ , then $λ$ is equal to _________ . [2023]

(36)

General term in the expansion of ${(\sqrt{x} - \frac{6}{x^{3 / 2}})}^{n}$ is given by
$T_{r + 1} = {}^{n}{C_{r}} {(x)}^{\frac{n - r}{2}} {(- 6)}^{r} {(x)}^{\frac{- 3}{2} r}$

For the constant term, the power of $x$ should be zero.

$\Rightarrow \frac{n - r}{2} - \frac{3}{2} r = 0 \Rightarrow n - 4 r = 0 \Rightarrow r = \frac{n}{4}$

For the constant term, put $x = 1$ in ${(\sqrt{x} - \frac{6}{x^{3 / 2}})}^{n}$

Constant term, ${(- 5)}^{n}$

$∴$ $Sum of coefficients except constant term is {(- 5)}^{n} - {}^{n}{C_{n / 4}} {(- 6)}^{n / 4} = 649$

Put $n = 4,$ $625 + 24 = 649$

Thus, $n = 4$ satisfies the above equation.

Now, $r = \frac{n}{4} = \frac{4}{4} = 1$

Required coefficient of $x^{- 4}$ = ${}^{4}{C_{3}} {(- 6)}^{3} = 4 (- 216)$

and constant term $a = {}^{4}{C_{1}} {(- 6)}^{1} = 4 (- 6)$

$∴$ $4 (- 216) = λ [4 (- 6)] \Rightarrow λ = 36$

Q 17 :

Let the sixth term in the binomial expansion of ${(\sqrt{2^{\log_{2} (10 - 3^{x})}} + \sqrt[5]{2^{(x - 2) \log_{2} 3}})}^{m}$ , in the increasing powers of $2^{(x - 2) \log_{2} 3}$ , be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $x$ is _______ . [2023]

(4)

Sixth term $T_{6} = {}^{m}{C_{5}} {(10 - 3^{x})}^{\frac{m - 5}{2}} \cdot (3^{x - 2}) = 21 ...(i)$

${}^{m}{C_{1}}, {}^{m}{C_{2}}, {}^{m}{C_{3}} are in A.P. \Rightarrow 2 {}^{m}{C_{2}} = {}^{m}{C_{1}} + {}^{m}{C_{3}}$

$\Rightarrow \frac{2 \cdot m!}{(m - 2)! 2!} = \frac{m!}{(m - 1)! 1!} + \frac{m!}{(m - 3)! 3!}$

$∴$ $m = 7, 2 (Rejected)$

Put $m = 7$ in equation (i):

${}^{7}{C_{5}} {(10 - 3^{x})}^{\frac{7 - 5}{2}} \cdot (3^{x - 2}) = 21$ $\Rightarrow 21 \cdot (10 - 3^{x}) \cdot \frac{3^{x}}{9} = 21$

$\Rightarrow x = 0, 2$

Sum of the squares of all possible values of $x$ is $0^{2} + 2^{2} = 4$

Q 18 :

If the term without $x$ in the expansion of ${(x^{\frac{2}{3}} + \frac{α}{x^{3}})}^{22}$ is 7315, then $| α |$ is equal to ________ . [2023]

(1)

$T_{r + 1} = {}^{22}C_{r} {(x^{\frac{2}{3}})}^{22 - r} \cdot {(\frac{α}{x^{3}})}^{r}$

$= {}^{22}C_{r} \cdot (x^{\frac{44}{3} - \frac{2 r}{3} - 3 r}) \cdot (α^{r})$

For the term without $' x';$ $\frac{44}{3} - \frac{2 r}{3} - 3 r = 0 \Rightarrow \frac{44}{3} = \frac{11 r}{3}$

$∴$ $r = 4$

So, the term without $x$ is ${}^{22}C_{4} α^{4} = 7315$

$\Rightarrow \frac{22 \times 21 \times 20 \times 19}{24} \cdot α^{4} = 7315$

$∴$ $| α |$ $= 1$

Q 19 :

The constant term in the expansion of ${(2 x + \frac{1}{x^{7}} + 3 x^{2})}^{5}$ is _______. [2023]

(1080)

$General term = \frac{5!}{r_{1}! r_{2}! r_{3}!} {(2 x)}^{r_{1}} {(x^{- 7})}^{r_{2}} {(3 x^{2})}^{r_{3}}$

$= \frac{5!}{r_{1}! r_{2}! r_{3}!} 2^{r_{1}} \cdot 3^{r_{3}} \cdot x^{r_{1} - 7 r_{2} + 2 r_{3}}$

For the constant term, $r_{1} - 7 r_{2} + 2 r_{3} = 0$ and $r_{1} + r_{2} + r_{3} = 5$

By hit and trial, $r_{1} = 1$ , $r_{2} = 1$ and $r_{3} = 3$ .

$∴$ $Constant term = \frac{5!}{1! 1! 3!} 2^{1} \cdot 3^{3}$

$= \frac{120}{6} \times 2 \times 27 = 20 \times 2 \times 27 = 1080$

Q 20 :

Let the coefficients of three consecutive terms in the binomial expansion of ${(1 + 2 x)}^{n}$ be in the ratio 2:5:8.Then the coefficient of the term, which is in the middle of these three terms, is _________ . [2023]

(1120)

$Coefficient of {(r + 1)}^{th} term is,$

$t_{r + 1} = {}^{n}C_{r} {(2 x)}^{r}$

Given consecutive term ratio as,

$\frac{{}^{n}C_{r - 1} {(2)}^{r - 1}}{{}^{n}C_{r} {(2)}^{r}} = \frac{2}{5} \Rightarrow \frac{n! / (r - 1)! (n - r + 1)!}{n! (2) / r! (n - r)!} = \frac{2}{5}$

$\Rightarrow \frac{r}{n - r + 1} = \frac{4}{5} \Rightarrow 5 r = 4 n - 4 r + 4 \Rightarrow 9 r = 4 (n + 1) ...(i)$

Again, consecutive term ratio are,

$\frac{{}^{n}C_{r} {(2)}^{r}}{{}^{n}C_{r + 1} {(2)}^{r + 1}} = \frac{5}{8} \Rightarrow \frac{n! / r! (n - r)!}{n! / (r + 1)! (n - r - 1)!} = \frac{5}{4}$

$\Rightarrow \frac{r + 1}{n - r} = \frac{5}{4} \Rightarrow 5 n - 4 = 9 r ...(ii)$

$From (i) and (ii): n = 8, r = 4$

$So, coefficient of middle term is,$

${}^{8}C_{4} 2^{4} = 16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 1120$

Q 21 :

If the co-efficient of $x^{9}$ in ${(α x^{3} + \frac{1}{β x})}^{11}$ and the co-efficient of $x^{- 9}$ in ${(α x - \frac{1}{β x^{3}})}^{11}$ are equal, then ${(α β)}^{2}$ is equal to ________ . [2023]

(1)

${(r + 1)}^{th} term in$ ${(α x^{3} + \frac{1}{β x})}^{11} = {}^{11}C_{r} α^{11 - r} \cdot β^{- r} \cdot x^{33 - 4 r}$

$Here, 33 - 4 r = 9 \Rightarrow r = 6$

$∴$ $Coefficient of x^{9} = {}^{11}C_{6} α^{5} β^{- 6}$

${(r + 1)}^{th} term in$ ${(α x - \frac{1}{β x^{3}})}^{11} = {}^{11}C_{r} α^{11 - r} {(- \frac{1}{β})}^{r} \cdot x^{11 - 4 r}$

$Here, 11 - 4 r = - 9 \Rightarrow r = 5$

$∴$ $Coefficient of x^{- 9} = - {}^{11}C_{5} α^{6} β^{- 5}$

$Now, {}^{11}C_{6} \frac{α^{5}}{β^{6}} = - {}^{11}C_{5} \frac{α^{6}}{β^{5}} [∵ Given] \Rightarrow \frac{1}{β} = - α \Rightarrow {(α β)}^{2} = 1$

Q 22 :

Let $α > 0$ , be the smallest number such that the expansion of ${(x^{\frac{2}{3}} + \frac{2}{x^{3}})}^{30}$ has a term $β x^{- α}, β \in N .$ Then $α$ is equal to ________ . [2023]

(2)

$Here, T_{r + 1} = {}^{30}C_{r} {(x^{2 / 3})}^{30 - r} {(\frac{2}{x^{3}})}^{r}$

$= {}^{30}C_{r} x^{(20 - \frac{2 r}{3} - 3 r)} {(2)}^{r} = {}^{30}C_{r} {(2)}^{r} x^{(\frac{60 - 11 r}{3})}; 0 \leq r \leq 30$

$For β x^{- α}, α > 0, r = 6, and T_{7} = {}^{30}C_{6} {(2)}^{6} x^{- 2} \Rightarrow α = 2$

Q 23 :

The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

(5040)

$T_{r + 1} = {}^{9}C_{r} {(\frac{4 x}{5})}^{9 - r} {(\frac{5}{2 x^{2}})}^{r}$

$= {}^{9}C_{r} {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} \cdot x^{9 - r - 2 r} \Rightarrow x^{9 - r - 2 r} = x^{- 6}$

$\Rightarrow 9 - r - 2 r = - 6 \Rightarrow 3 r = 15 \Rightarrow r = 5$

$So, T_{6} = {}^{9}C_{5} {(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5} x^{- 6}$

$= \frac{9!}{5! 4!} (\frac{4 \times 4 \times 4 \times 4}{5 \times 5 \times 5 \times 5}) (\frac{5 \times 5 \times 5 \times 5 \times 5}{2 \times 2 \times 2 \times 2 \times 2}) x^{- 6}$

$= \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2} (8) (5) x^{- 6} = 126 \times 40 \cdot x^{- 6} = 5040$

Q 24 :

If the constant term in the binomial expansion of ${(\frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^{l}})}^{9}$ is $- 84$ and the coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

(98)

$T_{r + 1} = {}^{9}C_{r} {\frac{(x^{5 / 2})}{2^{9 - r}}}^{9 - r} {(\frac{- 4}{x^{l}})}^{r}$

$= {(- 1)}^{r} \frac{{}^{9}C_{r}}{2^{9 - r}} 4^{r} x^{\frac{45}{2} - \frac{5 r}{2} - l r} \Rightarrow 45 - 5 r - 2 l r = 0$

$r = \frac{45}{5 + 2 l} \dots (i)$

$Now, according to the question, {(- 1)}^{r} \frac{{}^{9}C_{r}}{2^{9 - r}} 4^{r} = - 84$

$\Rightarrow {(- 1)}^{r} {}^{9}C_{r} 2^{3 r - 9} = - 21 \times 4$

$Only natural value of r possible if 3 r - 9 = 0 \Rightarrow r = 3 and {}^{9}C_{3} = 84$

$From equation (i): l = 5$

$For coefficient of x^{- 3 l} : \frac{45}{2} - \frac{5 r}{2} - l r = - 3 l \Rightarrow \frac{45}{2} - \frac{5 r}{2} - 5 r = - 3 l \Rightarrow r = 5$

$∴$ $The coefficient of x^{- 3 l} is {}^{9}C_{5} (- 1) \frac{4^{5}}{2^{4}} = 2^{α} \times β$

$\Rightarrow α = 7, β = - 63$

$∴$ $Value of$ $| α l - β |$ $= 7 \times 5 + 63 = 35 + 63 = 98$

Q 25 :

The sum of all rational terms in the expansion of ${(2^{\frac{1}{5}} + 5^{\frac{1}{3}})}^{15}$ is equal to [2024]

3133
633
931
6131

1

2

3

4

(1)

We have, ${(2^{1 / 5} + 5^{1 / 3})}^{15}$

$∴$ $T_{r + 1} = {}^{15}{C_{r}} {(2^{1 / 5})}^{15 - r} {(5^{1 / 3})}^{r} = {}^{15}{C_{r}} 5^{r / 3} 2^{(3 - \frac{r}{5})}$

For rational terms, $\frac{r}{3}$ and $\frac{r}{5}$ must be an integer

$∴$ 3 and 5 divide $r \Rightarrow 15$ divides $r \Rightarrow r = 0$ and $r = 15$

Hence, ${}^{15}{C_{0}} 5^{0} 2^{3} + {}^{15}{C_{15}} 5^{5} 2^{0}$

$∴$ Required sum = 8 + 3125 = 3133

Q 26 :

If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

639
742
724
693

1

2

3

4

(4)

$T_{r + 1} = {}^{12}{C_{r}} {(\frac{3^{1 / 5}}{x})}^{12 - r} {(\frac{2 x}{5^{1 / 3}})}^{r}$

$= {}^{12}C_{r} (\frac{3^{\frac{12 - r}{5}}}{5^{r / 3}}) \cdot 2^{r} \cdot x^{2 r - 12}$

For constant term, $2 r - 12 = 0$

$\Rightarrow r = 6$

$∴$ Constant term = $\frac{{}^{12}{C_{6} \cdot} 3^{6 / 5} \cdot 2^{6}}{5^{2}}$

$= 924 \times \frac{3^{1} 3^{1 / 5} \cdot 2^{6}}{25} = \frac{11 \times 9 \times 7 \times 2^{8} \times 3^{1 / 5}}{25}$

$\Rightarrow α = \frac{11 \times 9 \times 7}{25} \Rightarrow 25 α = 11 \times 9 \times 7 = 693$

Q 27 :

If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

2
6
4
9

1

2

3

4

(3)

We have, ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$

$T_{r + 1} = {}^{10}C_{r} {(\sqrt{a} x^{2})}^{10 - r} {(\frac{1}{2 x^{3}})}^{r}$

$= {}^{10}C_{r} {(\sqrt{a})}^{10 - r} {(\frac{1}{2})}^{r} x^{20 - 2 r - 3 r}$

For the term to be independent of $x,$ we have $20 - 2 r - 3 r = 0$

$\Rightarrow r = 4$

$∴$ Required term $= T_{5} = {}^{10}C_{4} {(\sqrt{a})}^{6} {(\frac{1}{2})}^{4} = 105 (given)$

$\Rightarrow \frac{210}{16} a^{3} = 105 \Rightarrow a^{3} = 8 \Rightarrow a = 2$

Hence, $a^{2} = 4$

Q 28 :

The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

$\frac{69}{16}$
$\frac{63}{16}$
$\frac{21}{4}$
$\frac{19}{4}$

1

2

3

4

(3)

$T_{r + 1} = {}^{9}{C_{r}} {(\frac{x^{- 2 / 5}}{2})}^{r} {(x^{2 / 3})}^{9 - r}$

$= {}^{9}{C_{r}} \frac{1}{2^{r}} x^{6 - \frac{2}{3} r - \frac{2}{5} r} = {}^{9}{C_{r}} \frac{1}{2^{r}} x^{6 - \frac{16 r}{15}}$

For coefficient of $x^{2 / 3}, 6 - \frac{16 r}{15} = \frac{2}{3}$

$\Rightarrow 90 - 16 r = 10$

$\Rightarrow r = 5$

For coefficient of $x^{- 2 / 5}, 6 - \frac{16 r}{15} = \frac{- 2}{5}$

$\Rightarrow 90 - 16 r = - 6$

$\Rightarrow r = 6$

Required sum $= {}^{9}{C_{5}} \frac{1}{2^{5}} + {}^{9}{C_{6}} \frac{1}{2^{6}} = \frac{126}{2^{5}} + \frac{84}{2^{6}}$

$= \frac{336}{2^{6}} = \frac{21}{4}$

Q 29 :

Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18} .$ Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is: [2024]

$\frac{1}{9}$
$\frac{1}{4}$
$\frac{4}{9}$
$\frac{9}{4}$

1

2

3

4

(4)

$T_{r + 1} = {}^{18}{C_{r}} {(\frac{1}{3} x^{1 / 3})}^{18 - r} \cdot {(\frac{1}{2} x^{- 2 / 3})}^{r}$

$∴$ Coefficient of $7^{th}$ term $= {}^{18}{C_{6}} {(\frac{1}{3})}^{12} \cdot {(\frac{1}{2})}^{6}$

and coefficient of $13^{th}$ term $= {}^{18}{C_{12}} {(\frac{1}{3})}^{6} \cdot {(\frac{1}{2})}^{12}$

So, ${(\frac{n}{m})}^{1 / 3} = {[\frac{{}^{18}{C_{12}} {(\frac{1}{3})}^{6} \cdot {(\frac{1}{2})}^{12}}{{}^{18}{C_{6}} {(\frac{1}{3})}^{12} \cdot {(\frac{1}{2})}^{6}}]}^{1 / 3} = \frac{9}{4}$

Q 30 :

If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]

(54)

General term of $I = {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is

$T_{r + 1} = {}^{9}{C_{r}} {(\frac{3}{2} x^{2})}^{9 - r} {(\frac{- 1}{3 x})}^{r}$

$= {}^{9}{C_{r}} {(\frac{3}{2})}^{9 - r} \frac{x^{18 - 2 r}}{x^{r}} {(\frac{- 1}{3})}^{r}$

$= {}^{9}{C_{r}} {(\frac{3}{2})}^{9 - r} r^{18 - 3 r} {(\frac{- 1}{3})}^{r}$

Now, $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ will have the constant term $= 1 \times coeff of x^{0} in I + 2 \times coeff of \frac{1}{x} in I - 3 \times coeff of x^{- 3} in I$

$= 1 \times {}^{9}{C_{6}} {(\frac{3}{2})}^{3} {(\frac{- 1}{3})}^{6} + 2 \times 0 - 3 \times {}^{9}{C_{7}} {(\frac{3}{2})}^{2} {(\frac{- 1}{3})}^{7}$

$= \frac{84}{2^{3} \times 3^{3}} + \frac{3 \times 36 \times 1}{2^{2} \times 3^{5}} = \frac{7}{2 \times 3^{2}} + \frac{2}{2 \times 3^{2}}$

$= \frac{9}{2 \times 3^{2}} = \frac{1}{2}$

$∴$ $p = \frac{1}{2} .$ So $108 p = \frac{1}{2} \times 108 = 54$

Topic Question Set

Q 1 :

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})}^{n}$ is $\sqrt{6} : 1$ , then the third term from the beginning is [2023]

(2)

Q 2 :

If the coefficients of $x^{7}$ in ${(a x^{2} + \frac{1}{2 b x})}^{11}$ and $x^{- 7}$ in ${(a x - \frac{1}{3 b x^{2}})}^{11}$ are equal, then [2023]

Q 3 :

If the coefficients of three consecutive terms in the expansion of ${(1 + x)}^{n}$ are in the ratio 1:5:20, then the coefficient of the fourth term is [2023]

Q 4 :

The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of ${(2 x^{2} + \frac{1}{2 x})}^{11}$ is equal to [2023]

Q 5 :

If the coefficient of $x^{7}$ in ${(a x - \frac{1}{b x^{2}})}^{13}$ and the coefficient of $x^{- 5}$ in ${(a x + \frac{1}{b x^{2}})}^{13}$ are equal, then $a^{4} b^{4}$ is equal to [2023]

Q 6 :

If the coefficients of $x$ and $x^{2}$ in ${(1 + x)}^{p} {(1 - x)}^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to [2023]

Q 7 :

The sum of the coefficients of three consecutive terms in the binomial expansion of ${(1 + x)}^{n + 2}$ , which are in the ratio 1:3:5, is equal to [2023]

Q 8 :

The coefficient of $x^{5}$ in the expansion of ${(2 x^{3} - \frac{1}{3 x^{2}})}^{5}$ is [2023]

Q 9 :

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of ${(1 + x)}^{10}$ , then $\sum_{r = 1}^{10} r^{3} {(\frac{a_{r}}{a_{r - 1}})}^{2}$ is equal to [2023]

(2)

Q 11 :

The coefficient of $x^{301}$ in ${(1 + x)}^{500} + x {(1 + x)}^{499} + x^{2} {(1 + x)}^{498} + . . . + x^{500}$ is [2023]

Q 12 :

If the coefficient of $x^{15}$ in the expansion of ${(a x^{3} + \frac{1}{b x^{1 / 3}})}^{15}$ is equal to the coefficient of $x^{- 15}$ in the expansion of ${(a x^{1 / 3} - \frac{1}{b x^{3}})}^{15}$ , where $a$ and $b$ are positive real numbers, then for each such ordered pair (a, b) [2023]

Q 13 :

The coefficient of $x^{18}$ in the expansion of ${(x^{4} - \frac{1}{x^{3}})}^{15}$ is __________ . [2023]

Q 14 :

If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]

Q 15 :

The number of integral terms in the expansion of ${(3^{1 / 2} + 5^{1 / 4})}^{680}$ is equal to ___________ . [2023]

Q 16 :

Let $α$ be the constant term in the binomial expansion of ${(\sqrt{x} - \frac{6}{x^{\frac{3}{2}}})}^{n}, n \leq 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{- n}$ is $λ α$ , then $λ$ is equal to _________ . [2023]

Q 18 :

If the term without $x$ in the expansion of ${(x^{\frac{2}{3}} + \frac{α}{x^{3}})}^{22}$ is 7315, then $| α |$ is equal to ________ . [2023]

Q 19 :

The constant term in the expansion of ${(2 x + \frac{1}{x^{7}} + 3 x^{2})}^{5}$ is _______. [2023]

Q 20 :

Let the coefficients of three consecutive terms in the binomial expansion of ${(1 + 2 x)}^{n}$ be in the ratio 2:5:8.Then the coefficient of the term, which is in the middle of these three terms, is _________ . [2023]

Q 21 :

If the co-efficient of $x^{9}$ in ${(α x^{3} + \frac{1}{β x})}^{11}$ and the co-efficient of $x^{- 9}$ in ${(α x - \frac{1}{β x^{3}})}^{11}$ are equal, then ${(α β)}^{2}$ is equal to ________ . [2023]

Q 22 :

Let $α > 0$ , be the smallest number such that the expansion of ${(x^{\frac{2}{3}} + \frac{2}{x^{3}})}^{30}$ has a term $β x^{- α}, β \in N .$ Then $α$ is equal to ________ . [2023]

Q 23 :

The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

Q 24 :

If the constant term in the binomial expansion of ${(\frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^{l}})}^{9}$ is $- 84$ and the coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

Q 25 :

The sum of all rational terms in the expansion of ${(2^{\frac{1}{5}} + 5^{\frac{1}{3}})}^{15}$ is equal to [2024]

Q 26 :

If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

Q 27 :

If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

Q 28 :

The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

Q 29 :

Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18} .$ Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is: [2024]

Q 30 :

If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]

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Topic Question Set

Q 1 : If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (24+134)n is 6:1, then the third term from the beginning is [2023]

(2)

Q 2 : If the coefficients of x7 in (ax2+12bx)11 and x-7 in (ax-13bx2)11 are equal, then [2023]

Q 3 : If the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1:5:20, then the coefficient of the fourth term is [2023]

Q 4 : The absolute difference of the coefficients of x10 and x7 in the expansion of (2x2+12x)11 is equal to [2023]

Q 5 : If the coefficient of x7 in (ax-1bx2)13 and the coefficient of x-5 in (ax+1bx2)13 are equal, then a4b4 is equal to [2023]

Q 6 : If the coefficients of x and x2 in (1+x)p(1-x)q are 4 and -5 respectively, then 2p+3q is equal to [2023]

Q 7 : The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2, which are in the ratio 1:3:5, is equal to [2023]

Q 8 : The coefficient of x5 in the expansion of (2x3-13x2)5 is [2023]

Q 9 : If ar is the coefficient of x10-r in the Binomial expansion of (1+x)10, then ∑r=110r3(arar-1)2 is equal to [2023]

(2)

Q 10 : Let K be the sum of the coefficients of the odd powers of x in the expansion of (1+x)99. Let a be the middle term in the expansion of (2+12)200. If C99K200a=2lmn, where m and n are odd numbers, then the ordered pair (l,n) is equal to [2023]

Q 11 : The coefficient of x301 in (1+x)500+x(1+x)499+x2(1+x)498+...+ x500 is [2023]

Q 12 : If the coefficient of x15 in the expansion of (ax3+1bx1/3)15 is equal to the coefficient of x-15 in the expansion of (ax1/3-1bx3)15, where a and b are positive real numbers, then for each such ordered pair (a, b) [2023]

Q 13 : The coefficient of x18 in the expansion of (x4-1x3)15 is __________ . [2023]

Q 14 : If the constant term in the expansion of (3x2-12x5)7 is α, then [α] is equal to _______ . [2023]

Q 15 : The number of integral terms in the expansion of (31/2+51/4)680 is equal to ___________ . [2023]

Q 16 : Let α be the constant term in the binomial expansion of (x-6x32)n,n≤15. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of x-n is λα, then λ is equal to _________ . [2023]

Q 18 : If the term without x in the expansion of (x23+αx3)22 is 7315, then |α| is equal to ________ . [2023]

Q 19 : The constant term in the expansion of (2x+1x7+3x2)5 is _______. [2023]

Q 20 : Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)n be in the ratio 2:5:8.Then the coefficient of the term, which is in the middle of these three terms, is _________ . [2023]

Q 21 : If the co-efficient of x9 in (αx3+1βx)11 and the co-efficient of x-9 in (αx-1βx3)11 are equal, then (αβ)2 is equal to ________ . [2023]

Q 22 : Let α>0, be the smallest number such that the expansion of (x23+2x3)30 has a term βx-α,β∈N. Then α is equal to ________ . [2023]

Q 23 : The coefficient of x-6, in the expansion of (4x5+52x2)9, is __________ . [2023]

Q 24 : If the constant term in the binomial expansion of (x522-4xl)9 is -84 and the coefficient of x-3l is 2αβ, where β<0 is an odd number, then |αl-β| is equal to __________ . [2023]

Q 25 : The sum of all rational terms in the expansion of (215+513)15 is equal to [2024]

Q 26 : If the constant term in the expansion of (35x+2x53)12, x≠0, is α×28×35, then 25α is equal to: [2024]

Q 27 : If the term independent of x in the expansion of (ax2+12x3)10 is 105, then a2 is equal to : [2024]

Q 28 : The sum of the coefficient of x23 and x-25 in the binomial expansion of (x23+12x-25)9 is [2024]

Q 29 : Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of (13x13+12x23)18. Then (nm)13 is: [2024]

Q 30 : If the constant term in the expansion of (1+2x-3x3)(32x2-13x)9 is p, then 108p is equal to _______. [2024]

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Q 1 :

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})}^{n}$ is $\sqrt{6} : 1$ , then the third term from the beginning is [2023]

Q 2 :

If the coefficients of $x^{7}$ in ${(a x^{2} + \frac{1}{2 b x})}^{11}$ and $x^{- 7}$ in ${(a x - \frac{1}{3 b x^{2}})}^{11}$ are equal, then [2023]

Q 3 :

If the coefficients of three consecutive terms in the expansion of ${(1 + x)}^{n}$ are in the ratio 1:5:20, then the coefficient of the fourth term is [2023]

Q 4 :

The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of ${(2 x^{2} + \frac{1}{2 x})}^{11}$ is equal to [2023]

Q 5 :

If the coefficient of $x^{7}$ in ${(a x - \frac{1}{b x^{2}})}^{13}$ and the coefficient of $x^{- 5}$ in ${(a x + \frac{1}{b x^{2}})}^{13}$ are equal, then $a^{4} b^{4}$ is equal to [2023]

Q 6 :

If the coefficients of $x$ and $x^{2}$ in ${(1 + x)}^{p} {(1 - x)}^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to [2023]

Q 7 :

The sum of the coefficients of three consecutive terms in the binomial expansion of ${(1 + x)}^{n + 2}$ , which are in the ratio 1:3:5, is equal to [2023]

Q 8 :

The coefficient of $x^{5}$ in the expansion of ${(2 x^{3} - \frac{1}{3 x^{2}})}^{5}$ is [2023]

Q 9 :

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of ${(1 + x)}^{10}$ , then $\sum_{r = 1}^{10} r^{3} {(\frac{a_{r}}{a_{r - 1}})}^{2}$ is equal to [2023]

Q 11 :

The coefficient of $x^{301}$ in ${(1 + x)}^{500} + x {(1 + x)}^{499} + x^{2} {(1 + x)}^{498} + . . . + x^{500}$ is [2023]

Q 12 :

If the coefficient of $x^{15}$ in the expansion of ${(a x^{3} + \frac{1}{b x^{1 / 3}})}^{15}$ is equal to the coefficient of $x^{- 15}$ in the expansion of ${(a x^{1 / 3} - \frac{1}{b x^{3}})}^{15}$ , where $a$ and $b$ are positive real numbers, then for each such ordered pair (a, b) [2023]

Q 13 :

The coefficient of $x^{18}$ in the expansion of ${(x^{4} - \frac{1}{x^{3}})}^{15}$ is __________ . [2023]

Q 14 :

If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]

Q 15 :

The number of integral terms in the expansion of ${(3^{1 / 2} + 5^{1 / 4})}^{680}$ is equal to ___________ . [2023]

Q 16 :

Let $α$ be the constant term in the binomial expansion of ${(\sqrt{x} - \frac{6}{x^{\frac{3}{2}}})}^{n}, n \leq 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{- n}$ is $λ α$ , then $λ$ is equal to _________ . [2023]

Q 18 :

If the term without $x$ in the expansion of ${(x^{\frac{2}{3}} + \frac{α}{x^{3}})}^{22}$ is 7315, then $| α |$ is equal to ________ . [2023]

Q 19 :

The constant term in the expansion of ${(2 x + \frac{1}{x^{7}} + 3 x^{2})}^{5}$ is _______. [2023]

Q 20 :

Let the coefficients of three consecutive terms in the binomial expansion of ${(1 + 2 x)}^{n}$ be in the ratio 2:5:8.Then the coefficient of the term, which is in the middle of these three terms, is _________ . [2023]

Q 21 :

If the co-efficient of $x^{9}$ in ${(α x^{3} + \frac{1}{β x})}^{11}$ and the co-efficient of $x^{- 9}$ in ${(α x - \frac{1}{β x^{3}})}^{11}$ are equal, then ${(α β)}^{2}$ is equal to ________ . [2023]

Q 22 :

Let $α > 0$ , be the smallest number such that the expansion of ${(x^{\frac{2}{3}} + \frac{2}{x^{3}})}^{30}$ has a term $β x^{- α}, β \in N .$ Then $α$ is equal to ________ . [2023]

Q 23 :

The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

Q 24 :

If the constant term in the binomial expansion of ${(\frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^{l}})}^{9}$ is $- 84$ and the coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

Q 25 :

The sum of all rational terms in the expansion of ${(2^{\frac{1}{5}} + 5^{\frac{1}{3}})}^{15}$ is equal to [2024]

Q 26 :

If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

Q 27 :

If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

Q 28 :

The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

Q 29 :

Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18} .$ Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is: [2024]

Q 30 :

If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]