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)

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

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

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

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

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

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

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

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

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)

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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]

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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]