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

Q 11 :

Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}, x^{5}, x^{3}$ ands x respectively in the expansion of ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}, x > 1$ . If u and v satisfy the equations $α u + β v = 18, γ u + δ v = 20$ , then u + v equals : [2025]

5
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8

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(1)

We have, ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}$

= $2 {{}^{5}C_{0} x^{5} + {}^{5}C_{2} x^{3} (x^{3} - 1) + {}^{5}C_{4} x {(x^{3} - 1)}^{2}}$

$∵ [{(x + a)}^{n} + {(x - a)}^{n} = 2 ({}^{n}C_{0} x^{n} + {}^{n}C_{2} x^{n - 2} a^{2} + . . .)]$

$= 2 {5 x^{7} + 10 x^{6} + x^{5} - 10 x^{4} - 10 x^{3} + 5 x}$

On comparing, we get

$α = 10, β = 2, γ = - 20, δ = 10$

Now, 10u + 2v = 18 and – 20u +10v = 20

$\Rightarrow$ u = 1, v = 4

Hence, u + v = 5.

Q 12 :

The remainder when $({{(64)}^{(64)})}^{(64)}$ is divided by 7 is equal to [2025]

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(2)

Let $A = ({{(64)}^{(64)})}^{(64)} = {(64)}^{64^{2}}$

$A = {(1 + 63)}^{64^{2}}$ , let $64^{2} = n$

Expanding by binomial

$A = {(1 + 63)}^{n} = 1 + {}^{n}C_{1} (63) + {}^{n}C_{2} {(63)}^{2} + . . . = 1 + 63 λ = 1 + 7 (9 λ)$

$∴$ When $({{(64)}^{(64)})}^{(64)}$ is divisible by 7 then remainder is 1.

Q 13 :

If in the expansion of ${(1 + x)}^{p} {(1 - x)}^{q}$ , the coefficients of x and $x^{2}$ are 1 and –2, respectively, then $p^{2} + q^{2}$ is equal to : [2025]

18
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(4)

${(1 + x)}^{p} {(1 - x)}^{q} = ({}^{p}C_{0} + C_{1} x + {}^{p}C_{2} x^{2} + . . .) ({}^{q}C_{0} - {}^{q}C_{1} x + {}^{p}C_{2} x^{2} + . . .)$

Coeff. of $x \equiv - {}^{p}C_{0} {}^{q}C_{1} + {}^{p}C_{1} {}^{q}C_{0} = 1$

$\Rightarrow p - q = 1$ ... (i)

Coeff. of $x^{2} \equiv {}^{p}C_{0} {}^{q}C_{2} - {}^{p}C_{1} {}^{q}C_{1} + {}^{p}C_{2} {}^{q}C_{0} = - 2$

$\Rightarrow \frac{q (q - 1)}{2} - p q + \frac{p (p - 1)}{2} = - 2$

$\Rightarrow q^{2} - q - 2 p q + p^{2} - p = - 4$

$\Rightarrow 1 - (p + q) = - 4$ $[∵ q^{2} + p^{2} - 2 p q = {(p - q)}^{2}]$

$\Rightarrow p + q = 5$ ... (ii)

On solving equation (i) and (ii), we get p = 3 and q = 2.

So, . $p^{2} + q^{2} = 13$

Q 14 :

Suppose A and B are the coefficients of $30^{th}$ and $12^{th}$ terms respectively in the binomial expansion of ${(1 + x^{2})}^{2 n - 1}$ . If 2A = 5B, then n is equal to: [2025]

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(3)

Given : A = Coefficient of $30^{th}$ term in ${(1 + x^{2})}^{2 n - 1}$

B = Coefficient of $12^{th}$ term in ${(1 + x^{2})}^{2 n - 1}$

They can also be written as

$A = {}^{2 n - 1}C_{29} and B = {}^{2 n - 1}C_{11}$

Since, 2A = 5B, then

$2 \cdot {}^{2 n - 1}C_{29} = 5 \cdot {}^{2 n - 1}C_{11}$

$\Rightarrow 2 \cdot \frac{(2 n - 1)!}{29! (2 n - 30)!} = 5 \cdot \frac{(2 n - 1)!}{11! (2 n - 12)!}$

$\Rightarrow \frac{1}{30 \times 29 \times 28 \times . . . \times 12}$

$= \frac{1}{(2 n - 12) (2 n - 13) (2 n - 14) . . . (2 n - 29) \times 12}$

On comparing, we get

2n – 12 = 30 $\Rightarrow$ n = 21.

Q 15 :

The remainder, when $7^{103}$ is divided by 23, is equal to: [2025]

6
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(2)

We have, $7^{103} = 7 (7^{102}) = 7 {(343)}^{34} = 7 {(345 - 2)}^{34}$

$= 7 {(- 2)}^{34} = 7 {(2)}^{34} = 7 \cdot 2^{2} \cdot 2^{32} = 28 \cdot {(256)}^{4} = 28 {(253 + 3)}^{4}$

$= 28 {(3)}^{4} = 28 \times 81 = (23 + 5) (69 + 12) = (5) (12) = 60$

$∴$ Remainder = 14.

Q 16 :

The sum of the series $2 \times 1 \times {}^{20}C_{4} - 3 \times 2 \times {}^{20}C_{5} + 4 \times 3 \times {}^{20}C_{6} - 5 \times 4 \times {}^{20}C_{7} + . . . + 18 \times 17 \times {}^{20}C_{20}$ , is equal to __________. [2025]

(34)

Let $2 \times 1 \times {}^{20}C_{4} - 3 \times 2 \times {}^{20}C_{5} + . . . + 18 \times 17 \times {}^{20}C_{20} = A$

Using binomial theorem,

${(1 + x)}^{20} = {}^{20}C_{0} + {}^{20}C_{1} x + {}^{20}C_{2} x^{2} + {}^{20}C_{3} x^{3} + . . . + {}^{20}C_{20} x^{20}$

$\frac{{(1 + x)}^{20}}{x^{2}} = \frac{1}{x^{2}} + \frac{20}{x} + {}^{20}C_{2} + {}^{20}C_{3} x + . . . + {}^{20}C_{20} x^{18}$

Differentiate with respect to x, we get

$\frac{2 x {(1 + x)}^{19} (9 x - 1)}{x^{4}} = \frac{- 2}{x^{3}} - \frac{20}{x^{2}} + {}^{20}C_{3} + 2 \cdot {}^{20}C_{4} x + . . . + 18 {}^{20}C_{20} x^{17}$

$\frac{2 {(1 + x)}^{19} (9 x - 1)}{x^{3}} = \frac{- 2}{x^{3}} - \frac{20}{x^{2}} + {}^{20}C_{3} + 2 \cdot {}^{20}C_{4} x + . . . + 18 {}^{20}C_{20} x^{17}$

Again, differentiate with respect to x, we get

$\frac{2 [x^{3} {(1 + x)}^{19} \cdot 9 + x^{3} (9 x - 1) \cdot 19 {(1 + x)}^{18} - {(1 + x)}^{19} (9 x - 1) \cdot 3 x^{2}]}{x^{6}}$

$= \frac{6}{x^{4}} + \frac{40}{x^{3}} + 2 \cdot 1 \cdot {}^{20}C_{4} + 3 \cdot 2 \cdot {}^{20}C_{5} x + . . . + 18 \cdot 17 \cdot {}^{20}C_{20} x^{16}$

Put x = –1 in above equation,

0 = 6 – 40 + A

$\Rightarrow$ A = 34.

Q 17 :

The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]

(63)

We have, ${(1919)}^{1919} = {(1920 - 1)}^{1919}$

$= {}^{1919}C_{0} {(1920)}^{1919} - {}^{1919}C_{1} {(1920)}^{1918} + . . . + {}^{1919}C_{1918} (1920) - {}^{1919}C_{1919}$

$= {(1920)}^{1919} - 1919 {(1920)}^{1918} + . . . + 1919 (1920) - 1$

$= 100 λ + 1919 (1920) - 1$

[As all terms will have 100 as multiple except the last two terms]

$= 100 λ + 3684480 - 1 = 100 λ + 3684479$

$∴$ Last two digits will be 79

So, required product = $7 \times 9 = 63$ .

Q 18 :

The sum of all possible values of $n \in N$ , so that the coefficients of $x, x^{2}$ and $x^{3}$ in the expansion of ${(1 + x^{2})}^{2} {(1 + x)}^{n}$ , are in arithmetic progression, is: [2026]

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(3)

$(x^{4} + 2 x^{2} + 1) ({}^{n}C_{0} x^{0} + {}^{n}C_{1} x^{1} + {}^{n}C_{2} x^{2} + {}^{n}C_{3} x^{3} + \dots)$

$Coefficient of x \Rightarrow {}^{n}C_{1}$

$Coeff. of x^{2} \Rightarrow 2 + {}^{n}C_{2} = 2 + \frac{n (n - 1)}{2}$

$Coeff. of x^{3} = 2 \cdot {}^{n}C_{1} + {}^{n}C_{3}$

$= 2 n + \frac{n (n - 1) (n - 2)}{6} (if x \geq 3)$

Now according to question

$n + 2 n + \frac{n (n - 1) (n - 2)}{6} = 2 [2 + \frac{n (n - 1)}{2}]$

$3 n + \frac{n (n - 1) (n - 2)}{6} = 4 + n (n - 1)$

$\Rightarrow n^{3} - 9 n^{2} + 26 n - 24 = 0$

$\Rightarrow n = 2, 3, 4 \Rightarrow n = 3, 4$

$Now checking for n = 2$

$\begin{matrix} Coeff. of x = 2 \\ Coeff. of x^{2} = 3 \\ Coeff. of x^{3} = 4 \end{matrix}} \Rightarrow are in A.P.$

$\Rightarrow n = 2 is also the correct choice$

$Required sum of values of n$

$= 2 + 3 + 4 = 9$

$Option (3)$

Q 19 :

The value of $\frac{{}^{100}C_{50}}{51} + \frac{{}^{100}C_{51}}{52} + \dots + \frac{{}^{100}C_{100}}{101}$ is: [2026]

$\frac{2^{100}}{101}$
$\frac{2^{100}}{100}$
$\frac{2^{101}}{101}$
$\frac{2^{101}}{100}$

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(1)

$S = \sum_{r = 50}^{100} \frac{{}^{100}{C_{r}}}{r + 1} = \sum_{r = 50}^{100} \frac{1}{r + 1} \cdot \frac{r + 1}{101} .^{101} C_{r + 1}$

$S = \frac{1}{101} \sum_{r = 50}^{100} {}^{101}C_{r + 1}$

$= \frac{1}{101} \times \frac{2^{101}}{2} = \frac{2^{100}}{101}$

$Option (1)$

Q 20 :

Let $S = \frac{1}{25!} + \frac{1}{3! 23!} + \frac{1}{5! 21!} + \dots$ up to 13 terms. If $13 S = \frac{2^{k}}{n!}, k \in ℕ,$ then $n + k$ is equal to [2026]

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(2)

$\frac{1}{26!} (\frac{26!}{25! 1!} + \frac{26!}{3! 23!} + \frac{26!}{5! 21!} + \dots + 13 terms)$

$= \frac{1}{26!} ({}^{26}C_{1} + {}^{26}C_{3} + {}^{26}C_{5} + \dots + 13 terms)$

$= \frac{1}{26!} ({}^{26}C_{1} + {}^{26}C_{3} + {}^{26}C_{5} + \dots + {}^{26}C_{25})$

$\Rightarrow S = \frac{1}{26!} \times 2^{25}$

$\Rightarrow 13 S = \frac{2^{24}}{25!}$

$so n + k = 25 + 24 = 49$

Q 21 :

The coefficient of $x^{48}$ in $(1 + x) + 2 {(1 + x)}^{2} + 3 {(1 + x)}^{3} + \dots + 100 {(1 + x)}^{100}$ is equal to [2026]

$100 \cdot {}^{100}{C_{49}} - {}^{100}{C_{48}}$
${}^{100}{C_{50}} + {}^{101}{C_{49}}$
$100 \cdot {}^{101}{C_{49}} - {}^{101}{C_{50}}$
$100 \cdot {}^{100}{C_{49}} - {}^{100}{C_{50}}$

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(3)

$Let 1 + x = r$

$∴$ $S = 1 \cdot r + 2 \cdot r^{2} + 3 \cdot r^{3} + \dots + 100 r^{100} . . . (1)$
$(AGP)$

$r S = 1 \cdot r^{2} + 2 \cdot r^{3} + \dots + 99 r^{100} + 100 r^{101} . . . (2)$

$(1) - (2) gives$

$S = - \frac{{(1 + x)}^{101}}{x^{2}} + \frac{1}{x^{2}} + \frac{100 {(1 + x)}^{101}}{x}$

$∴$ $coefficient of x^{48} in S$

$= - coefficient of x^{48} in \frac{{(1 + x)}^{101}}{x^{2}} + 100 . Coefficient of x^{48} in \frac{{(1 + x)}^{101}}{x}$

$= 100 . {}^{101}{C_{49}} - {}^{101}{C_{50}}$

Q 22 :

The sum of the coefficients of $x^{499} and x^{500}$ in ${(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$ is: [2026]

${}^{1002}C_{500}$
${}^{1000}C_{501}$
${}^{1001}C_{501}$
${}^{1002}C_{501}$

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(1)

$S = {(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$

$= {(1 + x)}^{1000} (\frac{1 - {(\frac{x}{1 + x})}^{1001}}{1 - \frac{x}{1 + x}})$

$= {(1 + x)}^{1001} - x^{1001}$

$Required sum = {}^{1001}{C_{499}} + {}^{1001}{C_{500}} = {}^{1002}{C_{500}}$

Q 23 :

Given below are two statements :

Statement I : $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by 7.

Statement II :The integral part of ${(7 + 4 \sqrt{3})}^{25}$ is an odd number.

In the light of the above statements, choose the correct answer from the options given below : [2026]

Both Statement I and Statement II are true
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Statement I is true but Statement II is false

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(1)

Statement I :

Statement II: $R = {(7 + 4 \sqrt{3})}^{25} = I + f$

$R' = {(7 - 4 \sqrt{3})}^{25} = f'$

$∴$ $R + R' = 2 [{}^{25}{C_{0}} 7^{25} + {}^{25}{C_{2}} 7^{23} {(4 \sqrt{3})}^{2} + \dots]$

$I + f + f' = even integer$

$∴$ $I = odd integer$

$∵$ $0 < f + f' < 2 \Rightarrow f + f' = 1$

$\Rightarrow Both the statements are correct$

Q 24 :

If $((\frac{1}{{}^{15}{C_{0}}} + \frac{1}{{}^{15}{C_{1}}})) ((\frac{1}{{}^{15}{C_{1}}} + \frac{1}{{}^{15}{C_{2}}})) \dots ((\frac{1}{{}^{15}{C_{12}}} + \frac{1}{{}^{15}{C_{13}}})) = \frac{α^{13}}{{}^{14}{C_{0}}^{14} C_{1} \dots^{14} C_{12}},$ then $30 α$ is equal to_____ [2026]

(32)

$\prod_{r = 0}^{12} (\frac{1}{{}^{15}{C_{r}}} + \frac{1}{{}^{15}{C_{r + 1}}}) = \prod_{r = 0}^{12} \frac{\frac{16}{r + 1} \cdot {}^{15}C_{r}}{15 C_{r} \cdot {}^{15}C_{r + 1}}$

$= \prod_{r = 0}^{12} \frac{16}{(r + 1) \cdot \frac{15}{r + 1} \cdot {}^{14}C_{r}} = \prod_{r = 0}^{12} \frac{\frac{16}{15}}{{}^{14}C_{r}}$

$= \frac{{(\frac{16}{15})}^{13}}{{}^{14}C_{0} \cdot {}^{14}C_{1} \dots {}^{14}C_{12}}$ $\Rightarrow α = \frac{16}{15}$

$\Rightarrow 30 α = 32$

Q 25 :

Let $C_{r}$ denote the coefficient of $x^{r}$ in the binomial expansion of $(1 + x)^{n}$ , $n \in ℕ$ $0 \leq r \leq n$ .

If $P_{n} = C_{0} - C_{1} + \frac{2^{2}}{3} C_{2} - \frac{2^{3}}{4} C_{3} + \dots + \frac{{(- 2)}^{n}}{n + 1} C_{n},$ then the value of $\sum_{n = 1}^{25} \frac{1}{P_{2 n}}$ equals: [2026]

675
525
650
580

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(1)

$P_{n} = \sum_{r = 0}^{n} \frac{{}^{n}C_{r} {(- 2)}^{r}}{r + 1} = \sum_{r = 0}^{n} \frac{1}{n + 1} {}^{n + 1}C_{r + 1} {(- 2)}^{r}$

$= \frac{- 1}{2 (n + 1)} \sum_{r = 0}^{n} {}^{n + 1}C_{r + 1} {(- 2)}^{r + 1}$

$= \frac{- 1}{2 (n + 1)} [{(1 - 2)}^{n + 1} - 1]$

$P_{n} = \frac{1}{2 (n + 1)} [1 - {(- 1)}^{n + 1}]$

$P_{2 n} = \frac{1}{2 (2 n + 1)} [1 - {(- 1)}^{2 n + 1}]$

$P_{2 n} = \frac{1}{2 n + 1}$

$\sum_{n = 1}^{25} \frac{1}{P_{2 n}} = \sum_{n = 1}^{25} (2 n + 1)$

$= 3 + 5 + \dots + 51$

$= \frac{25}{2} (51 + 3)$

$= 25 \times 27 = 675$

Q 26 :

Expand ${(x^{2} + \frac{3}{x})}^{4}, x \neq 0 .$

$x^{8} + 12 x^{5} + 54 x^{2} + \frac{108}{x} + \frac{81}{x^{4}}$
$x^{8} - 12 x^{5} + 54 x^{2} - \frac{108}{x} - \frac{81}{x^{4}}$
$x^{8} - 12 x^{5} - 54 x^{2} + \frac{108}{x} + \frac{81}{x^{4}}$
$None of these$

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(1)

$We have,$

${(x^{2} + \frac{3}{x})}^{4} = {}^{4}C_{0} {(x^{2})}^{4} + {}^{4}C_{1} {(x^{2})}^{3} (\frac{3}{x}) + {}^{4}C_{2} {(x^{2})}^{2} {(\frac{3}{x})}^{2}$ $+ {}^{4}C_{3} (x^{2}) {(\frac{3}{x})}^{3} + {}^{4}C_{4} {(\frac{3}{x})}^{4}$

$= x^{8} + 4 x^{6} \cdot \frac{3}{x} + 6 x^{4} \cdot \frac{9}{x^{2}} + 4 x^{2} \cdot \frac{27}{x^{3}} + \frac{81}{x^{4}}$

$= x^{8} + 12 x^{5} + 54 x^{2} + \frac{108}{x} + \frac{81}{x^{4}}$

Q 27 :

${(\frac{x}{3} + \frac{1}{x})}^{5} =$

$\frac{x^{5}}{243} + \frac{5 x^{3}}{81} + \frac{10 x}{27} + \frac{10}{9 x} + \frac{5}{3 x^{3}} + \frac{1}{x^{5}}$
$\frac{x^{5}}{420} + \frac{5 x^{4}}{2} + \frac{3 x^{3}}{81} + \frac{10 x}{27} + \frac{5}{x} + \frac{2}{x^{4}}$
$\frac{2 x^{5}}{243} + \frac{4 x^{4}}{3} + \frac{3 x^{3}}{81} + \frac{10 x}{9} + \frac{10}{9 x} + \frac{5}{3 x^{3}}$
None of these

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(1)

$We have, {(\frac{x}{3} + \frac{1}{x})}^{5}$

$= {}^{5}C_{0} {(\frac{x}{3})}^{5} + {}^{5}C_{1} {(\frac{x}{3})}^{4} (\frac{1}{x}) + {}^{5}C_{2} {(\frac{x}{3})}^{3} {(\frac{1}{x})}^{2}$ $+ {}^{5}C_{3} {(\frac{x}{3})}^{2} {(\frac{1}{x})}^{3} + {}^{5}C_{4} (\frac{x}{3}) {(\frac{1}{x})}^{4} + {}^{5}C_{5} {(\frac{1}{x})}^{5}$

$= \frac{x^{5}}{243} + \frac{5 x^{3}}{81} + \frac{10 x}{27} + \frac{10}{9 x} + \frac{5}{3 x^{3}} + \frac{1}{x^{5}}$

Q 28 :

The remainder when ${(32^{32})}^{32}$ is divided by 7, is

2
4
6
5

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(2)

$We have, {(32^{32})}^{32} = 32^{1024} = {(28 + 4)}^{1024}$

$= 4 + a multiple of 28 = 4 + a multiple of 7$

Q 29 :

Coefficient of $x$ in the expansion of ${(x^{2} + \frac{a}{x})}^{5}$ is

$9 a^{2}$
$10 a^{3}$
$10 a^{2}$
$10 a$

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(2)

$Given, {(x^{2} + \frac{a}{x})}^{5}$

$= {}^{5}C_{0} {(x^{2})}^{5} + {}^{5}C_{1} {(x^{2})}^{4} (\frac{a}{x}) + {}^{5}C_{2} {(x^{2})}^{3} {(\frac{a}{x})}^{2}$ $+ {}^{5}C_{3} {(x^{2})}^{2} {(\frac{a}{x})}^{3} + {}^{5}C_{4} x^{2} {(\frac{a}{x})}^{4} + {}^{5}C_{5} {(\frac{a}{x})}^{5}$

$∴$ $Coefficient of x = {}^{5}C_{3} a^{3} = 10 a^{3}$

Q 30 :

The coefficient of $x^{- 9}$ in the expansion of ${(\frac{x^{2}}{2} - \frac{2}{x})}^{9}$ is

512
- 512
521
251

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(2)

Q 31 :

The value of ${}^{14}C_{1} + {}^{14}C_{3} + {}^{14}C_{5} + \dots + {}^{14}C_{11}$ is

$2^{14} - 1$
$2^{14} - 14$
$2^{12}$
$2^{13} - 14$

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(4)

${}^{14}C_{1} + {}^{14}C_{3} + {}^{14}C_{5} + \dots + {}^{14}C_{11}$

$= ({}^{14}C_{1} + {}^{14}C_{3} + {}^{14}C_{5} + \dots + {}^{14}C_{13}) - {}^{14}C_{13}$

$= 2^{14 - 1} - {}^{14}C_{1} = 2^{13} - 14$

Q 32 :

Which of the following results is valid?

${(1 + x)}^{n} > (1 + n x) for all natural numbers n$
${(1 + x)}^{n} \geq (1 + n x) for all natural numbers n, where x > - 1$
${(1 + x)}^{n} \leq (1 + n x) for all natural numbers n$
${(1 + x)}^{n} < (1 + n x) for all natural numbers n$

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(2)

Q 33 :

The value of ${{}^{15}C}_{0}^{2} - {{}^{15}C}_{1}^{2} + {{}^{15}C}_{2}^{2} - \dots - {{}^{15}C}_{15}^{2}$ is

0
- 15
15
51

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(1)

$Let f (n) = {{}^{n}C}_{0}^{2} - {{}^{n}C}_{1}^{2} + {{}^{n}C}_{2}^{2} - {{}^{n}C}_{3}^{2} + \dots + {(- 1)}^{n} {{}^{n}C}_{n}^{2}$

$∵$ $f (n) =$ ${\begin{matrix} 0, & n = odd \\ {(- 1)}^{n / 2}, {}^{n}C_{n / 2}, & n = even \end{matrix}$

Here, $n = 15$ (odd)

$∴$ ${{}^{15}C}_{0}^{2} - {{}^{15}C}_{1}^{2} + {{}^{15}C}_{2}^{2} - \dots - {{}^{15}C}_{15}^{2} = 0$

Q 34 :

If $s_{n} = \sum_{r = 0}^{n} \frac{1}{{}^{n}C_{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{{}^{n}C_{r}},$ then $\frac{t_{n}}{s_{n}}$ is equal to

$n - 1$
$\frac{1}{2} n - 1$
$\frac{1}{2} n$
$\frac{2 n - 1}{2}$

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(3)

$t_{n} = \sum_{r = 0}^{n} \frac{r}{{}^{n}C_{r}} = \sum_{r = 0}^{n} \frac{n - (n - r)}{{}^{n}C_{n - r}}$

$\Rightarrow t_{n} = n \sum_{r = 0}^{n} \frac{1}{{}^{n}C_{r}} - \sum_{r = 0}^{n} \frac{n - r}{{}^{n}C_{n - r}}$

$t_{n} = n \sum_{r = 0}^{n} \frac{1}{{}^{n}C_{r}} - \sum_{r = 0}^{n} \frac{r}{{}^{n}C_{r}}$ $[Replacing n - r by r]$

$t_{n} = n s_{n} - t_{n}$ $∴$ $\frac{t_{n}}{s_{n}} = \frac{n}{2}$

Q 35 :

In the expansion of ${(1 + 3 x + 2 x^{2})}^{6}$ , the coefficient of $x^{11}$ is

144
288
216
576

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(4)

$We have, {(1 + 3 x + 2 x^{2})}^{6} = ((1 + 3 x) + {(2 x^{2}))}^{6}$

$= {}^{6}C_{0} {(1 + 3 x)}^{6} + {}^{6}C_{1} {(1 + 3 x)}^{5} 2 x^{2} + {}^{6}C_{2} {(1 + 3 x)}^{4} {(2 x^{2})}^{2}$ $+ {}^{6}C_{3} {(1 + 3 x)}^{3} {(2 x^{2})}^{3} + {}^{6}C_{4} {(1 + 3 x)}^{2} {(2 x^{2})}^{4}$ $+ {}^{6}C_{5} (1 + 3 x) {(2 x^{2})}^{5} + {}^{6}C_{6} {(2 x^{2})}^{6}$

Coefficient of $x^{11}$ in above expression is

${}^{6}C_{5} \cdot 3 \cdot 32 = 6 \times 3 \times 32 = 576$

Q 36 :

$\sum_{k = 0}^{n} \frac{{}^{n}C_{k}}{k + 1} =$

$\frac{2^{n} - 1}{n + 1}$
$\frac{2^{n - 1} - 1}{n - 1}$
$\frac{2^{n + 1} - 1}{n + 1}$
$\frac{2^{n} - 1}{n}$

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4

(3)

$\sum_{k = 0}^{n} \frac{{}^{n}C_{k}}{k + 1} = \sum_{k = 0}^{n} \frac{n!}{(k + 1) k! (n - k)!}$

$= \sum_{k = 0}^{n} \frac{n! (n + 1)}{(k + 1)! (n - k)! (n + 1)}$

$= \frac{1}{n + 1} \sum_{k = 0}^{n} \frac{(n + 1)!}{(k + 1)! (n + 1 - k - 1)!}$

$= \frac{1}{n + 1} \sum_{k = 0}^{n} {}^{n + 1}C_{k + 1} = \frac{1}{n + 1} [2^{n + 1} - 1]$

Q 37 :

If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} C_{1} + C_{1} C_{2} + \dots + C_{n - 1} C_{n}$ is

$\frac{(2 n)!}{(n - 1)! (n + 1)!}$
$\frac{n!}{(n - 1)! (n + 1)!}$
$\frac{(n - 1)!}{n! (n + 1)!}$
none of these

1

2

3

4

(1)

${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ ...(i)

and ${(1 + \frac{1}{x})}^{n} = C_{0} + C_{1} \frac{1}{x} + C_{2} \frac{1}{x^{2}} + \dots + C_{n - 1} \frac{1}{x^{n - 1}} + C_{n} \frac{1}{x^{n}}$ ...(ii)

Multiplying (i) and (ii) and equating the coefficient of $x$ in $\frac{1}{x^{n}} {(1 + x)}^{2 n},$

we get $C_{0} C_{1} + C_{1} C_{2} + C_{2} C_{3} + \dots + C_{n - 1} C_{n} = {}^{2 n}C_{n + 1}$

$= \frac{(2 n)!}{(2 n - n - 1)! (n + 1)!} = \frac{(2 n)!}{(n - 1)! (n + 1)!}$

Q 38 :

If $A = {4^{n} - 3 n - 1 : n \in N}$ and $B = {9 (n - 1) : n \in N},$ then

$B \subset A$
$A \cup B = N$
$A \subset B$
none of these

1

2

3

4

(3)

Q 39 :

The sum of the coefficients in the binomial expansion of ${(\frac{1}{x} + 2 x)}^{6}$ is equal to

1024
729
243
512

1

2

3

4

(2)

${(\frac{1}{x} + 2 x)}^{6} = {}^{6}C_{0} {(\frac{1}{x})}^{6} {(2 x)}^{0} + {}^{6}C_{1} {(\frac{1}{x})}^{5} {(2 x)}^{1}$ $+ {}^{6}C_{2} {(\frac{1}{x})}^{4} {(2 x)}^{2} + {}^{6}C_{3} {(\frac{1}{x})}^{3} {(2 x)}^{3} + {}^{6}C_{4} {(\frac{1}{x})}^{2} {(2 x)}^{4}$ $+ {}^{6}C_{5} (\frac{1}{x}) {(2 x)}^{5} + {}^{6}C_{6} {(\frac{1}{x})}^{0} {(2 x)}^{6}$

$= {(\frac{1}{x})}^{6} + 12 {(\frac{1}{x})}^{4} + 60 {(\frac{1}{x})}^{2} + 160 + 240 x^{2} + 192 x^{4} + 64 x^{6}$

$Sum of coefficients = 1 + 12 + 60 + 160 + 240 + 192 + 64 = 729$

Q 40 :

If the coefficient of $x^{7}$ in ${[a x^{2} + (\frac{1}{b x})]}^{11}$ equals the coefficient of $x^{- 7}$ in ${[a x - (\frac{1}{b x^{2}})]}^{11},$ then $a$ and $b$ satisfy the relation

$a + b = 1$
$a - b = 1$
$a b = 1$
$\frac{a}{b} = 1$

1

2

3

4

(3)

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

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

$= {}^{11}C_{r} \frac{a^{11 - r}}{b^{r}} x^{11 - 3 r} {(- 1)}^{r}$

$∴$ $Coefficient of x^{7} in$

${(a x^{2} + \frac{1}{b x})}^{11} = {}^{11}C_{5} \frac{a^{6}}{b^{5}}$

$and coefficient of x^{- 7} in$

${(a x - \frac{1}{b x^{2}})}^{11} = {}^{11}C_{6} \frac{a^{5}}{b^{6}}$

$Now, {}^{11}C_{5} \frac{a^{6}}{b^{5}} = {}^{11}C_{6} \frac{a^{5}}{b^{6}}$

$∴$ $a b = 1$

Topic Question Set

Q 11 :

Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}, x^{5}, x^{3}$ ands x respectively in the expansion of ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}, x > 1$ . If u and v satisfy the equations $α u + β v = 18, γ u + δ v = 20$ , then u + v equals : [2025]

Q 12 :

The remainder when $({{(64)}^{(64)})}^{(64)}$ is divided by 7 is equal to [2025]

Q 13 :

If in the expansion of ${(1 + x)}^{p} {(1 - x)}^{q}$ , the coefficients of x and $x^{2}$ are 1 and –2, respectively, then $p^{2} + q^{2}$ is equal to : [2025]

Q 14 :

Suppose A and B are the coefficients of $30^{th}$ and $12^{th}$ terms respectively in the binomial expansion of ${(1 + x^{2})}^{2 n - 1}$ . If 2A = 5B, then n is equal to: [2025]

Q 15 :

The remainder, when $7^{103}$ is divided by 23, is equal to: [2025]

Q 16 :

The sum of the series $2 \times 1 \times {}^{20}C_{4} - 3 \times 2 \times {}^{20}C_{5} + 4 \times 3 \times {}^{20}C_{6} - 5 \times 4 \times {}^{20}C_{7} + . . . + 18 \times 17 \times {}^{20}C_{20}$ , is equal to __________. [2025]

Q 17 :

The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]

Q 18 :

The sum of all possible values of $n \in N$ , so that the coefficients of $x, x^{2}$ and $x^{3}$ in the expansion of ${(1 + x^{2})}^{2} {(1 + x)}^{n}$ , are in arithmetic progression, is: [2026]

Q 19 :

The value of $\frac{{}^{100}C_{50}}{51} + \frac{{}^{100}C_{51}}{52} + \dots + \frac{{}^{100}C_{100}}{101}$ is: [2026]

Q 20 :

Let $S = \frac{1}{25!} + \frac{1}{3! 23!} + \frac{1}{5! 21!} + \dots$ up to 13 terms. If $13 S = \frac{2^{k}}{n!}, k \in ℕ,$ then $n + k$ is equal to [2026]

Q 21 :

The coefficient of $x^{48}$ in $(1 + x) + 2 {(1 + x)}^{2} + 3 {(1 + x)}^{3} + \dots + 100 {(1 + x)}^{100}$ is equal to [2026]

Q 22 :

The sum of the coefficients of $x^{499} and x^{500}$ in ${(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$ is: [2026]

Q 23 :

Given below are two statements :

Statement I : $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by 7.

Statement II :The integral part of ${(7 + 4 \sqrt{3})}^{25}$ is an odd number.

In the light of the above statements, choose the correct answer from the options given below : [2026]

Q 24 :

If $((\frac{1}{{}^{15}{C_{0}}} + \frac{1}{{}^{15}{C_{1}}})) ((\frac{1}{{}^{15}{C_{1}}} + \frac{1}{{}^{15}{C_{2}}})) \dots ((\frac{1}{{}^{15}{C_{12}}} + \frac{1}{{}^{15}{C_{13}}})) = \frac{α^{13}}{{}^{14}{C_{0}}^{14} C_{1} \dots^{14} C_{12}},$ then $30 α$ is equal to_____ [2026]

Q 26 :

Expand ${(x^{2} + \frac{3}{x})}^{4}, x \neq 0 .$

Q 27 :

${(\frac{x}{3} + \frac{1}{x})}^{5} =$

Q 28 :

The remainder when ${(32^{32})}^{32}$ is divided by 7, is

Q 29 :

Coefficient of $x$ in the expansion of ${(x^{2} + \frac{a}{x})}^{5}$ is

Q 30 :

The coefficient of $x^{- 9}$ in the expansion of ${(\frac{x^{2}}{2} - \frac{2}{x})}^{9}$ is

Q 31 :

The value of ${}^{14}C_{1} + {}^{14}C_{3} + {}^{14}C_{5} + \dots + {}^{14}C_{11}$ is

Q 32 :

Which of the following results is valid?

Q 33 :

The value of ${{}^{15}C}_{0}^{2} - {{}^{15}C}_{1}^{2} + {{}^{15}C}_{2}^{2} - \dots - {{}^{15}C}_{15}^{2}$ is

Q 34 :

If $s_{n} = \sum_{r = 0}^{n} \frac{1}{{}^{n}C_{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{{}^{n}C_{r}},$ then $\frac{t_{n}}{s_{n}}$ is equal to

Q 35 :

In the expansion of ${(1 + 3 x + 2 x^{2})}^{6}$ , the coefficient of $x^{11}$ is

Q 36 :

$\sum_{k = 0}^{n} \frac{{}^{n}C_{k}}{k + 1} =$

Q 37 :

If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} C_{1} + C_{1} C_{2} + \dots + C_{n - 1} C_{n}$ is

Q 38 :

If $A = {4^{n} - 3 n - 1 : n \in N}$ and $B = {9 (n - 1) : n \in N},$ then

Q 39 :

The sum of the coefficients in the binomial expansion of ${(\frac{1}{x} + 2 x)}^{6}$ is equal to

Q 40 :

If the coefficient of $x^{7}$ in ${[a x^{2} + (\frac{1}{b x})]}^{11}$ equals the coefficient of $x^{- 7}$ in ${[a x - (\frac{1}{b x^{2}})]}^{11},$ then $a$ and $b$ satisfy the relation

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Topic Question Set

Q 11 : Let α,β,γ and δ be the coefficients of x7,x5,x3 ands x respectively in the expansion of (x+x3–1)5+(x–x3–1)5, x>1. If u and v satisfy the equations αu+βv=18, γu+δv=20, then u + v equals : [2025]

Q 12 : The remainder when ((64)(64))(64) is divided by 7 is equal to [2025]

Q 13 : If in the expansion of (1+x)p(1–x)q, the coefficients of x and x2 are 1 and –2, respectively, then p2+q2 is equal to : [2025]

Q 14 : Suppose A and B are the coefficients of 30th and 12th terms respectively in the binomial expansion of (1+x2)2n–1. If 2A = 5B, then n is equal to: [2025]

Q 15 : The remainder, when 7103 is divided by 23, is equal to: [2025]

Q 16 : The sum of the series 2×1×C420–3×2×C520+4×3×C620–5×4×C720+...+18×17×C2020, is equal to __________. [2025]

Q 17 : The product of the last two digits of (1919)1919 is __________. [2025]

Q 18 : The sum of all possible values of n∈N, so that the coefficients of x,x2 and x3 in the expansion of (1+x2)2(1+x)n, are in arithmetic progression, is: [2026]

Q 19 : The value of C5010051+C5110052+⋯+C100100101 is: [2026]

Q 20 : Let S=125!+13! 23!+15! 21!+⋯up to 13 terms. If 13S=2kn!, k∈ℕ, then n+k is equal to [2026]

Q 21 : The coefficient of x48 in (1+x)+2(1+x)2+3(1+x)3+⋯+100(1+x)100 is equal to [2026]

Q 22 : The sum of the coefficients of x499 and x500 in (1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000 is: [2026]

Q 23 : Given below are two statements : Statement I : 2513+2013+813+313 is divisible by 7. Statement II :The integral part of (7+43)25 is an odd number. In the light of the above statements, choose the correct answer from the options given below : [2026]

Q 24 : If (1C015+1C115)(1C115+1C215)⋯(1C1215+1C1315)=α13C014 14C1⋯ 14C12, then 30α is equal to_____ [2026]

Q 25 : Let Cr denote the coefficient of xr in the binomial expansion of (1+x)n, n∈ℕ 0≤r≤n. If Pn=C0-C1+223 C2-234C3+⋯+(-2)nn+1Cn, then the value of ∑n=1251P2n equals: [2026]

Q 26 : Expand (x2+3x)4, x≠0.

Q 27 : (x3+1x)5=

Q 28 : The remainder when (3232)32 is divided by 7, is

Q 29 : Coefficient of x in the expansion of (x2+ax)5 is

Q 30 : The coefficient of x-9 in the expansion of (x22-2x)9 is

Q 31 : The value of C114+C314+C514+⋯+C1114 is

Q 32 : Which of the following results is valid?

Q 33 : The value of C1502-C1512+C1522-⋯-C15152 is

Q 34 : If sn=∑r=0n1Crn and tn=∑r=0nrCrn, then tnsn is equal to

Q 35 : In the expansion of (1+3x+2x2)6, the coefficient of x11 is

Q 36 : ∑k=0nCknk+1=

Q 37 : If (1+x)n=C0+C1x+C2x2+⋯+Cnxn, then C0C1+C1C2+⋯+Cn-1Cn is

Q 38 : If A={4n-3n-1:n∈N} and B={9(n-1):n∈N}, then

Q 39 : The sum of the coefficients in the binomial expansion of (1x+2x)6 is equal to

Q 40 : If the coefficient of x7 in [ax2+(1bx)]11 equals the coefficient of x-7 in [ax-(1bx2)]11, then a and b satisfy the relation

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Q 11 :

Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}, x^{5}, x^{3}$ ands x respectively in the expansion of ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}, x > 1$ . If u and v satisfy the equations $α u + β v = 18, γ u + δ v = 20$ , then u + v equals : [2025]

Q 12 :

The remainder when $({{(64)}^{(64)})}^{(64)}$ is divided by 7 is equal to [2025]

Q 13 :

If in the expansion of ${(1 + x)}^{p} {(1 - x)}^{q}$ , the coefficients of x and $x^{2}$ are 1 and –2, respectively, then $p^{2} + q^{2}$ is equal to : [2025]

Q 14 :

Suppose A and B are the coefficients of $30^{th}$ and $12^{th}$ terms respectively in the binomial expansion of ${(1 + x^{2})}^{2 n - 1}$ . If 2A = 5B, then n is equal to: [2025]

Q 15 :

The remainder, when $7^{103}$ is divided by 23, is equal to: [2025]

Q 16 :

The sum of the series $2 \times 1 \times {}^{20}C_{4} - 3 \times 2 \times {}^{20}C_{5} + 4 \times 3 \times {}^{20}C_{6} - 5 \times 4 \times {}^{20}C_{7} + . . . + 18 \times 17 \times {}^{20}C_{20}$ , is equal to __________. [2025]

Q 17 :

The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]

Q 18 :

The sum of all possible values of $n \in N$ , so that the coefficients of $x, x^{2}$ and $x^{3}$ in the expansion of ${(1 + x^{2})}^{2} {(1 + x)}^{n}$ , are in arithmetic progression, is: [2026]

Q 19 :

The value of $\frac{{}^{100}C_{50}}{51} + \frac{{}^{100}C_{51}}{52} + \dots + \frac{{}^{100}C_{100}}{101}$ is: [2026]

Q 20 :

Let $S = \frac{1}{25!} + \frac{1}{3! 23!} + \frac{1}{5! 21!} + \dots$ up to 13 terms. If $13 S = \frac{2^{k}}{n!}, k \in ℕ,$ then $n + k$ is equal to [2026]

Q 21 :

The coefficient of $x^{48}$ in $(1 + x) + 2 {(1 + x)}^{2} + 3 {(1 + x)}^{3} + \dots + 100 {(1 + x)}^{100}$ is equal to [2026]

Q 22 :

The sum of the coefficients of $x^{499} and x^{500}$ in ${(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$ is: [2026]

Q 23 :

Given below are two statements :

Statement I : $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by 7.

Statement II :The integral part of ${(7 + 4 \sqrt{3})}^{25}$ is an odd number.

In the light of the above statements, choose the correct answer from the options given below : [2026]

Q 24 :

If $((\frac{1}{{}^{15}{C_{0}}} + \frac{1}{{}^{15}{C_{1}}})) ((\frac{1}{{}^{15}{C_{1}}} + \frac{1}{{}^{15}{C_{2}}})) \dots ((\frac{1}{{}^{15}{C_{12}}} + \frac{1}{{}^{15}{C_{13}}})) = \frac{α^{13}}{{}^{14}{C_{0}}^{14} C_{1} \dots^{14} C_{12}},$ then $30 α$ is equal to_____ [2026]

Q 26 :

Expand ${(x^{2} + \frac{3}{x})}^{4}, x \neq 0 .$

Q 27 :

${(\frac{x}{3} + \frac{1}{x})}^{5} =$

Q 28 :

The remainder when ${(32^{32})}^{32}$ is divided by 7, is

Q 29 :

Coefficient of $x$ in the expansion of ${(x^{2} + \frac{a}{x})}^{5}$ is

Q 30 :

The coefficient of $x^{- 9}$ in the expansion of ${(\frac{x^{2}}{2} - \frac{2}{x})}^{9}$ is

Q 31 :

The value of ${}^{14}C_{1} + {}^{14}C_{3} + {}^{14}C_{5} + \dots + {}^{14}C_{11}$ is

Q 32 :

Which of the following results is valid?

Q 33 :

The value of ${{}^{15}C}_{0}^{2} - {{}^{15}C}_{1}^{2} + {{}^{15}C}_{2}^{2} - \dots - {{}^{15}C}_{15}^{2}$ is

Q 34 :

If $s_{n} = \sum_{r = 0}^{n} \frac{1}{{}^{n}C_{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{{}^{n}C_{r}},$ then $\frac{t_{n}}{s_{n}}$ is equal to

Q 35 :

In the expansion of ${(1 + 3 x + 2 x^{2})}^{6}$ , the coefficient of $x^{11}$ is

Q 36 :

$\sum_{k = 0}^{n} \frac{{}^{n}C_{k}}{k + 1} =$

Q 37 :

If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} C_{1} + C_{1} C_{2} + \dots + C_{n - 1} C_{n}$ is

Q 38 :

If $A = {4^{n} - 3 n - 1 : n \in N}$ and $B = {9 (n - 1) : n \in N},$ then

Q 39 :

The sum of the coefficients in the binomial expansion of ${(\frac{1}{x} + 2 x)}^{6}$ is equal to

Q 40 :

If the coefficient of $x^{7}$ in ${[a x^{2} + (\frac{1}{b x})]}^{11}$ equals the coefficient of $x^{- 7}$ in ${[a x - (\frac{1}{b x^{2}})]}^{11},$ then $a$ and $b$ satisfy the relation