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

Q 1 :
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

(A)

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 2 :
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

(D)

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

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

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 3 :
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

(C)

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 4 :
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

(C)

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

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

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 5 :
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

(D)

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

$∴$ 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 6 :
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]

0%

(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 $108 p = \frac{1}{2} \times 108 = 54$

Q 7 :
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

0%

(806)

$T_{r + 1} = {}^{n}C_{r} x^{n - r} y^{r}$

$∴$ ${}^{n}{C_{1}} x^{n - 1} y = 135,$ ...(i)

${}^{n}{C_{2}} x^{n - 2} y^{2} = 30$ ...(ii)

and ${}^{n}{C_{3}} x^{n - 3} y^{3} = \frac{10}{3}$ ...(iii)

By (i) and (ii), we have

$\frac{{}^{n}{C_{1}} x}{{}^{n}{C_{2}} y} = \frac{135}{30} = \frac{9}{2}$ ...(iv)

By (ii) and (iii), we have

$\frac{{}^{n}{C_{2}} x}{{}^{n}{C_{3}} y} = \frac{30}{\frac{10}{3}} = 9$ ...(v)

By (iv) and (v), we have

$\frac{{}^{n}{C_{1}} x}{{}^{n}{C_{2}} y} \times \frac{{}^{n}{C_{3}} y}{{}^{n}{C_{2}} x} = \frac{\frac{9}{2}}{9}$

$\Rightarrow \frac{{}^{n}{C_{1} {}^{n}{C_{3}}}}{{({}^{n}{C_{2}})}^{2}} = \frac{1}{2} \Rightarrow 2 {}^{n}{C_{1}} {}^{n}{C_{3}} = {({}^{n}{C_{2}})}^{2}$

$\Rightarrow \frac{2 n n!}{3! (n - 3)!} = {(\frac{n!}{2! (n - 2)!})}^{2}$

$\Rightarrow \frac{2 n^{2} (n - 1) (n - 2)}{6} = {(\frac{n (n - 1)}{2})}^{2}$

$\Rightarrow \frac{n^{2} (n - 1) (n - 2)}{3} = \frac{n^{2} {(n - 1)}^{2}}{4} \Rightarrow \frac{n - 2}{3} = \frac{n - 1}{4}$

$\Rightarrow 4 n - 8 = 3 n - 3 \Rightarrow n = 5$

$\Rightarrow \frac{{}^{5}{C_{2} x}}{{}^{5}{C_{3}} y} = 9$ [Using (v)]

$\Rightarrow \frac{x}{y} = 9$

$\Rightarrow x = 9 y$ ...(vi)

$∴$ From (i), ${}^{5}{C_{1}} x^{4} \frac{x}{9} = 135 \Rightarrow x^{5} = \frac{135 \times 9}{5}$

$\Rightarrow x^{5} = 3^{5}$

$\Rightarrow x = 3$ and $y = \frac{1}{3}$ [Using (vi)]

Hence, $6 (n^{3} + x^{2} + y) = 6 (125 + 9 + \frac{1}{3}) = 806$

Q 8 :
The coefficient of $x^{2012}$ in the expansion of ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$ is equal to _______ . [2024]

0%

(0)

We have, ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$

$= {(1 - x)}^{2007} (1 - x) {(1 + x + x^{2})}^{2007} = (1 - x) {(1 - x^{3})}^{2007}$

$\Rightarrow (1 - x) ({}^{2007}{C_{0} - {}^{2007}{C_{1}} (x^{3}) + . . .})$

General term is $(1 - x) ({(- 1)}^{r} {}^{2007}{C_{r}} x^{3 r})$

$= {(- 1)}^{r} {}^{2007}{C_{r}} x^{3 r} - {(- 1)}^{r} {}^{2007}{C_{r}} x^{3 r + 1}$

Now, $3 r = 2012 \Rightarrow r \neq \frac{2012}{3}$

So, $3 r + 1 = 2012 \Rightarrow 3 r = 2011 \Rightarrow r \neq \frac{2011}{3}$

Hence, there is no term containing $x^{2012} .$

So, coefficient of $x^{2012} = 0$

Q 9 :
Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$ is equal to _______ . [2024]

0%

(138)

We have, ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$

Now, ${(n + 1)}^{th}$ term $= t_{n + 1} = {}^{n}{C_{r}} 11^{\frac{r}{6}} 7^{\frac{824 - r}{2}}$

For integral term, 6 should divide $r$ and $\frac{824 - r}{2}$ must be integer.

$\Rightarrow 2$ must divide $r$ and $r$ is divisible by 6

Possible values of $r$ so that integral terms are obtained i.e., 0, 6, 12, ... 822

Now, this is an A.P. so let $n$ be number of integral terms

$∴$ $822 = 0 + (n - 1) 6 \Rightarrow n = 138$

So, 138 integral terms will be there in the expansion.

Q 10 :
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]

0%

(118)

The given expansion $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}$

$\Rightarrow {(1 + x)}^{2} (1 - x) \frac{{(x^{3} + 3 x^{2} + 3 x + 1)}^{5}}{x^{15}}$

$\Rightarrow \frac{{(1 + x)}^{2} (1 - x) [{(x + 1)}^{3}]^{5}}{x^{15}} \Rightarrow \frac{(1 - x) {(1 + x)}^{17}}{x^{15}}$

$∴$ Coefficient of $x^{3} \Rightarrow x^{18} in (1 - x) {(1 + x)}^{17}$

$\Rightarrow {(1 + x)}^{17} - x {(1 + x)}^{17}$

$\Rightarrow 0 - x ({}^{17}{C_{17}} x^{17}) = {}^{17}{C_{17}} (- 1) = - 1$

and coefficient of $x^{- 13} \Rightarrow x^{2} in (1 - x) {(1 + x)}^{17}$

$\Rightarrow {(1 + x)}^{17} - x {(1 + x)}^{17}$

$\Rightarrow {}^{17}{C_{2}} - {}^{17}{C_{1}} = 17 \times 8 - 17 = 17 \times 7 = 119$

The sum of the coefficient $x^{3}$ and $x^{- 13} = - 1 + 119 = 118$

Topic Question Set

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

Q 2 :
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 3 :
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 4 :
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 5 :
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 6 :
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]

0%

Q 7 :
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

0%

Q 8 :
The coefficient of $x^{2012}$ in the expansion of ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$ is equal to _______ . [2024]

0%

Q 9 :
Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$ is equal to _______ . [2024]

0%

Q 10 :
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]

0%

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Topic Question Set

Q 1 : The sum of all rational terms in the expansion of (215+513)15 is equal to [2024]

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

Q 3 : If the term independent of x in the expansion of (ax2+12x3)10 is 105, then a2 is equal to : [2024]

Q 4 : The sum of the coefficient of x23 and x-25 in the binomial expansion of (x23+12x-25)9 is [2024]

Q 5 : 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 6 : If the constant term in the expansion of (1+2x-3x3)(32x2-13x)9 is p, then 108p is equal to _______. [2024]

0%

Q 7 : If the second, third and fourth terms in the expansion of (x+y)n are 135, 30, and 103, respectively, then 6(n3+x2+y) is equal to ______. [2024]

0%

Q 8 : The coefficient of x2012 in the expansion of (1-x)2008(1+x+x2)2007 is equal to _______ . [2024]

0%

Q 9 : Number of integral terms in the expansion of {7(12)+11(16)}824 is equal to _______ . [2024]

0%

Q 10 : In the expansion of (1+x)(1-x2)(1+3x+3x2+1x3)5,x≠0, the sum of the coefficients of x3 and x-13 is equal to _______ . [2024]

0%

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

Q 2 :
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 3 :
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 4 :
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 5 :
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 6 :
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]

Q 7 :
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

Q 8 :
The coefficient of $x^{2012}$ in the expansion of ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$ is equal to _______ . [2024]

Q 9 :
Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$ is equal to _______ . [2024]

Q 10 :
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]