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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(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]

3
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4

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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]

22
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20

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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
14
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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)

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

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

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

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

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

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

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

$= 3 + 5 + \dots + 51$

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

$= 25 \times 27 = 675$

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]

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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]

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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]