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

Q 1 :

Among the statements

$(S 1) : 2023^{2022} - 1999^{2022} i s d i v i s i b l e b y 8$

$(S 2) : 13 {(13)}^{n} - 11 n - 13 i s d i v i s i b l e b y 144 f o r i n f i n i t e l y m a n y n \in ℕ$ [2023]

only (S2) is correct
both (S1) and (S2) are incorrect
both (S1) and (S2) are correct
only (S1) is correct

1

2

3

4

(3)

As we know that $x^{n} - y^{n}$ is always divisible by $x - y$

$∴$ $2023^{2022} - 1999^{2022}$ is divisible by 2023 - 1999 = 24, which is divisible by 8. $∴$ $S 1 is true.$

Also, $13 {(13)}^{n} - 11 n - 13 = 13 {(12 + 1)}^{n} - 11 n - 13$

$= 13 [{}^{n}{C_{0}} 12^{n} + {}^{n}{C_{1}} 12^{n - 1} + {}^{n}{C_{2}} 12^{n - 2} + \dots + {}^{n}{C_{n}}] - 11 n - 13$

$= 13 [{}^{n}{C_{0}} 12^{n} + {}^{n}{C_{1}} 12^{n - 1} + \dots + {}^{n}{C_{n - 2}} 12^{2}] + 13 \cdot 12 \cdot n + 13 - 11 n - 13$

$= 13 \times 12^{2} [{}^{n}{C_{0}} 12^{n - 2} + {}^{n}{C_{1}} 12^{n - 3} + \dots + {}^{n}{C_{n - 2}}] + 145 n$

Which is divisible by 144, for all $n \geq 144$ .

$i.e., It is divisible by 144 for infinitely many n \in N .$

$∴$ $S 2 is also true.$

Q 2 :

$25^{190} - 19^{190} - 8^{190} + 2^{190}$ is divisible by [2023]

both 14 and 34
34 but not by 14
14 but not by 34
neither 14 nor 34

1

2

3

4

(2)

Given, $25^{190} - 19^{190} - 8^{190} + 2^{190}$

So, ${(17 + 8)}^{190} - 8^{190} + {(19 - 17)}^{190} - 19^{190}$

$= {}^{190}{C_{0}} 17^{190} + {}^{190}{C_{1}} 17^{189} \times 8 + \dots + 8^{190} - 8^{190} + {}^{190}{C_{0}} 19^{190} - {}^{190}{C_{1}} 19^{189} \times 17 + \dots + 17^{190} - 19^{190}$

$= 17 (2 K_{1}) - 17 (2 K_{2}), where K_{1} and K_{2} are integers.$

So, given number is divisible by 34 but not by 14.

Q 3 :

Let the number ${(22)}^{2022} + {(2022)}^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ is equal to [2023]

20
13
5
10

1

2

3

4

(3)

We are, ${(22)}^{2022} + {(2022)}^{22} = {(21 + 1)}^{2022} + {(2022)}^{22}$

Here ${(2022)}^{22}$ is divisible by 3 as 2022 is divisible by 3.

Expanding ${(21 + 1)}^{2022}$ , we get

${(21 + 1)}^{2022} = {}^{2022}{C_{0}} {(21)}^{2022} + {}^{2022}{C_{1}} {(21)}^{2021} + \dots + {}^{2022}{C_{2022}} {(1)}^{2022}$

$= [{(3)}^{2022} {(7)}^{2022} + {}^{2022}{C_{1}} {(3)}^{2021} 7^{2021} + \dots] + 1$

$= 3 [{(3)}^{2021} {(7)}^{2022} + {}^{2022}{C_{1}} {(3)}^{2020} {(7)}^{2021} + \dots] + 1$

$= 3 m + 1 ∴ Remainder, α = 1$

Also, ${(21 + 1)}^{22} + {(2023 - 1)}^{22}$

$= 7 [{(3)}^{2022} {(7)}^{2021} + {}^{2022}{C_{1}} {(3)}^{2021} {(7)}^{2020} + \dots] + 1 + {}^{22}{C_{0}} {(2023)}^{22} - {}^{22}{C_{1}} {(2023)}^{21} + \dots - {}^{22}C_{21} (2023) + {}^{22}{C_{22}}$

$= (7 r_{1} + 1) + 7 [(7^{21}) {(289)}^{22} - {}^{22}{C_{1}} {(7)}^{20} {(289)}^{21} + \dots - {}^{22}{C_{21}} (289)] + 1$

$= (7 r_{1} + 1) + (7 r_{2} + 1) = 7 (r_{1} + r_{2}) + 2 = 7 n + 2$

$∴ Remainder, β = 2$

Now, $α^{2} + β^{2} = 1^{2} + 2^{2} = 5$

Q 4 :

If $\frac{1}{n + 1} {}^{n}{C_{n}} + \frac{1}{n} {}^{n}{C_{n - 1}} + . . . . + \frac{1}{2} {}^{n}{C_{1}} + {}^{n}{C_{0}} = \frac{1023}{10}, t h e n n i s e q u a l t o$ [2023]

8
9
6
7

1

2

3

4

(2)

$\frac{1}{n + 1} {}^{n}{C_{n}} + \frac{1}{n} {}^{n}{C_{n - 1}} + \dots + {}^{n}{C_{0}}$

$= \sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1} = \frac{1}{n + 1} \sum_{r = 0}^{n} {}^{n + 1}{C_{r + 1}} = \frac{1}{n + 1} (2^{n + 1} - 1) = \frac{1023}{10} = \frac{2^{10} - 1}{10}$

$So, n + 1 = 10 \Rightarrow n = 9$

Q 5 :

The sum of the coefficients of the first 50 terms in the binomial expansion of ${(1 - x)}^{100},$ is equal to [2023]

${}^{101}{C_{50}}$
$- {}^{99}{C_{49}}$
$- {}^{101}{C_{50}}$
${}^{99}{C_{49}}$

1

2

3

4

(2)

${(1 - x)}^{100} = {}^{100}{C_{0}} - {}^{100}{C_{1}} x + {}^{100}{C_{2}} x^{2} - {}^{100}{C_{3}} x^{3} + \dots + {}^{100}{C_{100}} x^{100}$

or
${(1 - x)}^{100} = C_{0} - C_{1} x + C_{2} x^{2} - C_{3} x^{3} + \dots + C_{100} x^{100}$

If $x$ = 1, then

$0 = C_{0} - C_{1} + C_{2} - C_{3} + \dots - C_{99} + C_{100}$

$\Rightarrow 0 = 2 (C_{0} - C_{1} + C_{2} - \dots - C_{49}) + C_{50}$

$\Rightarrow C_{0} - C_{1} + C_{2} - \dots - C_{49} = \frac{- C_{50}}{2} = \frac{- 1}{2} \frac{100!}{50! 50!} = \frac{- 99!}{50! 49!}$

$\Rightarrow C_{0} - C_{1} + C_{2} - \dots - C_{49} = - {}^{99}{C_{49}}$

Q 6 :

Fractional part of the number $\frac{4^{2022}}{15}$ is equal to [2023]

$\frac{1}{15}$
$\frac{4}{15}$
$\frac{14}{15}$
$\frac{8}{15}$

1

2

3

4

(1)

$We have,$ ${\frac{4^{2022}}{15}}$ $= {\frac{{(4^{2})}^{1011}}{15}}$ $= {\frac{{(1 + 15)}^{1011}}{15}}$

$= {\frac{1}{15} + \frac{15^{1010}}{15} + \dots}$ $= {\frac{1}{15} + 0 + \dots}$ $= {\frac{1}{15}}$

Q 7 :

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + . . . + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is [2023]

$\frac{2^{51}}{50!}$
$\frac{2^{50}}{50!}$
$\frac{2^{51}}{51!}$
$\frac{2^{50}}{51!}$

1

2

3

4

(4)

$\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + \dots + \frac{1}{49! 2!} + \frac{1}{51! 1!}$

$= \frac{1}{51!} [\frac{51!}{1! 50!} + \frac{51!}{3! 48!} + \frac{51!}{5! 46!} + \dots + \frac{51!}{49! 2!} + \frac{51!}{51! 1!}]$

$= \frac{1}{51!} [{}^{51}{C_{1}} + {}^{51}{C_{3}} + \dots + {}^{51}{C_{51}}]$

$= \frac{1}{51!} 2^{51 - 1} = \frac{1}{51!} 2^{50}$

Q 8 :

The value of $\sum_{r = 0}^{22} {}^{22}{C_{r} {}^{23}{C_{r}}}$ is [2023]

${}^{45}{C_{24}}$
${}^{45}{C_{23}}$
${}^{44}{C_{23}}$
${}^{44}{C_{22}}$

1

2

3

4

(2)

$\sum_{r = 0}^{22} {}^{22}{C_{r}} {}^{23}{C_{r}}$

$= \sum_{r = 0}^{22} {}^{22}{C_{r}} {}^{23}{C_{23 - r}}$

$= {}^{22 + 23}{C_{23}} = {}^{45}{C_{23}}$

Q 9 :

If ${({}^{30}{C_{1}})}^{2} + 2 {({}^{30}{C_{2}})}^{2} + 3 {({}^{30}{C_{3}})}^{2} + . . . + 30 {({}^{30}{C_{30}})}^{2} = \frac{α 60!}{{(30!)}^{2}},$ then $α$ is equal to [2023]

60
10
15
30

1

2

3

4

(3)

$Given,$ ${({}^{30}C_{1})}^{2} + 2 {({}^{30}C_{2})}^{2} + 3 {({}^{30}C_{3})}^{2} + \dots + 30 {({}^{30}C_{30})}^{2} = \frac{α 60!}{{(30!)}^{2}}$

$\sum_{r = 1}^{30} r \cdot {({}^{30}C_{r})}^{2}$ $= \sum_{r = 1}^{30} r \cdot {}^{30}C_{r} \cdot {}^{30}C_{r}$

$= \sum_{r = 1}^{30} r \cdot \frac{30}{r} \cdot {}^{29}C_{r - 1} \cdot {}^{30}C_{r}$ $= \sum_{r = 1}^{30} 30 \cdot {}^{29}C_{r - 1} {}^{30}C_{30 - r}$ $= 30 {}^{59}C_{30}$

$\Rightarrow 30 \frac{59!}{30! 29!} \frac{30}{30} = \frac{15 \cdot 60!}{{(30)}^{2}} ∴ α = 15$

Q 10 :

Let $x = {(8 \sqrt{3} + 13)}^{13}$ and $y = {(7 \sqrt{2} + 9)}^{9} .$ If $[t]$ denotes the greatest integer $\leq t$ , then [2023]

[x] is even but [y] is odd
[x] + [y] is even
[x] and [y] are both odd
[x] is odd but [y] is even

1

2

3

4

(2)

$x = {(8 \sqrt{3} + 13)}^{13}$ $= {}^{13}C_{0} \cdot {(8 \sqrt{3})}^{13} + {}^{13}C_{1} {(8 \sqrt{3})}^{12} {(13)}^{1} + \dots$

And $x' = {(8 \sqrt{3} - 13)}^{13}$ $= {}^{13}C_{0} {(8 \sqrt{3})}^{13} - {}^{13}C_{1} {(8 \sqrt{3})}^{12} {(13)}^{1} + \dots$

$∴$ $x - x' = 2 [{}^{13}C_{1} \cdot {(8 \sqrt{3})}^{12} {(13)}^{1} + {}^{13}C_{3} {(8 \sqrt{3})}^{10} \cdot {(13)}^{3} + \dots]$

Thus, $x - x'$ is an even integer, hence $[x]$ is even.

Now, $y = {(7 \sqrt{2} + 9)}^{9} = {}^{9}C_{0} {(7 \sqrt{2})}^{9} + {}^{9}C_{1} {(7 \sqrt{2})}^{8} {(9)}^{1} + {}^{9}C_{2} {(7 \sqrt{2})}^{7} {(9)}^{2} + \dots$

And $y' = {(7 \sqrt{2} - 9)}^{9} = {}^{9}C_{0} {(7 \sqrt{2})}^{9} - {}^{9}C_{1} {(7 \sqrt{2})}^{8} {(9)}^{1} + {}^{9}C_{2} {(7 \sqrt{2})}^{7} {(9)}^{2} \dots$

$∴$ $y - y' = 2 [{}^{9}C_{1} {(7 \sqrt{2})}^{8} {(9)}^{1} + {}^{9}C_{3} {(7 \sqrt{2})}^{6} {(9)}^{3} + \dots]$

Thus, $y - y'$ is an even integer, hence $[y]$ is even.

Q 11 :

The largest natural number n such that $3^{n}$ divides 66! is _______ . [2023]

(31)

The largest prime $p$ in $n!$ is

$E_{p} (n!) = [\frac{n}{p}] + [\frac{n}{p^{2}}] + [\frac{n}{p^{3}}] + \dots$

Here, $p$ = 3 is a prime number and $n$ = 66

$∴$ $E_{3} (66!) = [\frac{66}{3}] + [\frac{66}{9}] + [\frac{66}{27}] + [\frac{66}{81}] + \dots$

$= 22 + 7 + 2 + 0 + \dots = 31$ $∴$ $66! = 3^{31} \dots$

$∴ The maximum exponent of 3 is 31.$

Q 12 :

The remainder, when $7^{103}$ is divided by 17, is ________ . [2023]

(12)

$7^{103} = 7 . 7^{102}$

$7 . 7^{102} = 7 {(7^{2})}^{51} = 7 {(49)}^{51} = 7 {(51 - 2)}^{51}$

$= 7 {(- 2)}^{51} = - 7 {(2)}^{51} = - 7 (2^{48 + 3}) = - 7.8 (2^{48})$

$= - 56 {(2^{4})}^{12} = - 56 {(17 - 1)}^{12} = - 56 \times {(- 1)}^{12} = (- 56)$

$Remainder = 12$

Q 13 :

The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

(29)

$23^{200} + 19^{200} = {(21 + 2)}^{200} + {(21 - 2)}^{200}$

$= 2 [{}^{200}{C_{0}} 21^{200} 2^{0} + {}^{200}{C_{2}} 21^{198} 2^{2} + \dots + {}^{200}{C_{200}} 21^{0} 2^{200}]$ ...(i)

Now, $21^{2}$ will be divisible by 49, so (i) can be written as $2 [49 λ + 2^{200}]$

$\frac{2 [49 λ + 2^{200}]}{49}$ $then, Remainder = \frac{2^{201}}{49} = \frac{{(8)}^{67}}{49} = \frac{{(7 + 1)}^{67}}{49}$

$= \frac{{}^{67}{C_{0}} 7^{67} \cdot 1^{0} + {}^{67}{C_{1}} 7^{66} \cdot 1 + \dots + {}^{67}{C_{66}} 7 \cdot 1^{66} + {}^{67}{C_{67} \cdot 1}}{49}$

$Remainder = \frac{67 \times 7 + 1}{49}$ $[other terms will be divisible by 49]$

$= \frac{(49 + 18) \times 7 + 1}{49}, Remainder = 29$

Q 14 :

Suppose $\sum_{r = 0}^{2023} r^{2} {}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ___________ . [2023]

(1012)

$\sum_{r = 0}^{2023} r^{2}$ ${}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$

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

Then, $\sum_{r = 0}^{2023} r^{2} \cdot {}^{2023}{C_{r}} = (2023) (2023 + 1) \cdot 2^{2023 - 2}$

$= 2023 \times 2024 \times 2^{2021} = 2023 \times α \times 2^{2022}$

$∴$ $2 α = 2024 \Rightarrow α = 1012$

Q 15 :

Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

(405)

Sum of coefficients of first three terms of ${(x - \frac{3}{x^{2}})}^{n}$ is 376.

${}^{n}C_{0} - {}^{n}C_{1} \cdot 3 + {}^{n}C_{2} \cdot 3^{2} = 376 \Rightarrow 1 - 3 n + \frac{n (n - 1)}{2} \times 9 = 376$

$\Rightarrow 3 n^{2} - 5 n - 250 = 0 ∴ n = 10$

Now, $T_{r + 1} = {}^{10}C_{r} x^{10 - r} {(\frac{- 3}{x^{2}})}^{r} = {}^{10}C_{r} {(- 3)}^{r} \cdot x^{10 - 3 r}$

For coefficient of $x^{4}$ $\Rightarrow 10 - 3 r = 4 \Rightarrow r = 2$

Coefficient of $x^{4}$ $= {}^{10}C_{2} {(- 3)}^{2} = 405$

Q 16 :

The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

(7)

${(2023)}^{2023} = {(2030 - 7)}^{2023} = {(35 K - 7)}^{2023}$

$= {}^{2023}{C_{0}} {(35 K)}^{2023} {(- 7)}^{0} + {}^{2023}{C_{1}} {(35 K)}^{2022} {(- 7)}^{1} + \dots + {}^{2023}{C_{2023}} {(- 7)}^{2023}$

$= 35 N - 7^{2023}$

Now, $- 7^{2023} = - 7 \times 7^{2022} = - 7 {(7^{2})}^{1011} = - 7 {(50 - 1)}^{1011}$

$= - 7 [{}^{1011}{C_{0} (50)}^{1011} - {}^{1011}{C_{1}} {(50)}^{1010} + \dots + {}^{1011}{C_{1011}}]$

$= - 7 (5 λ - 1) = 35 λ + 7$

$∴ When {(2023)}^{2023} is divided by 35, remainder is 7 .$

Q 17 :

50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

(23)

Given, $x^{1 / 50} = 12 \Rightarrow x = {(12)}^{50}$

$y^{1 / 50} = 18 \Rightarrow y = {(18)}^{50}$

$∴$ $\frac{x + y}{25} = remainder \Rightarrow \frac{x + y}{25} = \frac{12^{50} + 18^{50}}{25}$

$\frac{12^{50}}{25} + \frac{{(18)}^{50}}{25} = \frac{{(144)}^{25} + {(324)}^{25}}{25} = \frac{{(150 - 6)}^{25} + {(325 - 1)}^{25}}{25}$

$= \frac{25 k - (6^{25} + 1)}{25} = \frac{25 k - [{(5 + 1)}^{25} + 1]}{25} = \frac{25 k_{1} - 2}{25}$

$∴$ $Remainder = 23$

Q 18 :

The remainder on dividing $5^{99}$ by 11 is _________ . [2023]

(9)

$We have, 5^{99} = 5^{4} \times 5^{95} = 625 \times {(5^{5})}^{19}$

$= 625 \times {(3125)}^{19} = 625 \times {(3124 + 1)}^{19}$

$= 625 \times {(11 λ + 1)}^{19} [∵ 11 \times 284 = 3124]$

$∴$ $Remainder = 9$

Q 19 :

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of ${(1 + x)}^{n}$ are in the arithmetic progression, then the maximum value of $n$ is: [2024]

14
21
7
28

1

2

3

4

(1)

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

$∴$ ${}^{n}{C_{5}} - {}^{n}{C_{4}} = {}^{n}{C_{6}} - {}^{n}{C_{5}} [Since, {}^{n}{C_{4}}, {}^{n}{C_{5}} and {}^{n}{C_{6}} are in A.P.]$

$\Rightarrow \frac{n!}{5! (n - 5)!} - \frac{n!}{4! (n - 4)!} = \frac{n!}{6! (n - 6)!} - \frac{n!}{5! (n - 5)!}$

$\Rightarrow \frac{n - 9}{5! (n - 5)! (n - 4)} = \frac{n - 11}{6! (n - 6)! (n - 5)}$

$\Rightarrow 6 (n - 9) = (n - 11) (n - 4)$

$\Rightarrow n^{2} - 21 n + 98 = 0 \Rightarrow n = \frac{21 \pm \sqrt{441 - 392}}{2} = 14, 7$

$∴$ $n_{max} = 14$

Q 20 :

The coefficient of $x^{70}$ in $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + x^{4} {(1 + x)}^{96} + \dots + x^{54} {(1 + x)}^{46}$ is ${}^{99}{C_{p} - {}^{46}{C_{q} .}}$ Then a possible value of $p + q$ is: [2024]

68
83
55
61

1

2

3

4

(2)

We have, $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + \dots + x^{54} {(1 + x)}^{46}$

It is a G.P. with first term $= x^{2} {(1 + x)}^{98}$

and common ratio $= \frac{x}{1 + x}$

$∴$ Sum of these terms $= x^{2} {(1 + x)}^{98} (\frac{{(\frac{x}{1 + x})}^{53} - 1}{\frac{x}{1 + x} - 1})$

$= x^{2} {(1 + x)}^{98} ((1 + x) - x^{53} {(1 + x)}^{- 52})$

$= x^{2} \underset{coeff . of x^{68}}{\underset{⏝}{{(1 + x)}^{99}}}$ $- x^{55}$ $\underset{coeff . of x^{15}}{\underset{⏝}{{(1 + x)}^{46}}}$ [ $∵$ we need coefficient of $x^{70}$ ]

$= {}^{99}{C_{68}} - {}^{46}{C_{15}} = {}^{99}{C_{p}} - {}^{46}{C_{q}}$ ( $∵$ Given)

Hence, $p + q = 83$

Q 21 :

Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

(5)

$α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$

$= 4 \sum_{r = 1}^{n} r^{2} \frac{n}{r} {}^{n - 1}{C_{r - 1}} + 2 \sum_{r = 1}^{n} r \frac{n}{r} {}^{n - 1}{C_{r - 1}} + \sum_{r = 0}^{n} {}^{n}{C_{r}}$

$= 4 n \sum_{r = 1}^{n} r \cdot {}^{n - 1}{C_{r - 1}} + 2 n \sum_{r = 1}^{n} {}^{n - 1}{C_{r - 1}} + \sum_{r = 0}^{n} {}^{n}{C_{r}}$

$= 4 n (n + 1) 2^{n - 2} + 2 n 2^{n - 1} + 2^{n} = 2^{n} [n (n + 1) + n + 1]$

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

Now, $β = \sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1} + \frac{1}{n + 1} = \sum_{r = 0}^{n} \frac{{}^{n + 1}{C_{r + 1}}}{n + 1} + \frac{1}{n + 1}$

$= \frac{1}{n + 1} \cdot 2^{n + 1} \Rightarrow \frac{2 α}{β} = 2 \cdot \frac{2^{n} {(n + 1)}^{2} \times (n + 1)}{2^{n + 1}} = {(n + 1)}^{3}$

Given that, $140 < \frac{2 α}{β} < 281 \Rightarrow 140 < {(n + 1)}^{3} < 281$

Now, for $n = 4; {(n + 1)}^{3} = 5^{3} = 125$

for $n = 5; {(n + 1)}^{3} = 6^{3} = 216$

for $n = 6; {(n + 1)}^{3} = 7^{3} = 343$

Hence, the value of $n = 5$

Q 22 :

The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

(1)

$428^{2024} = {(420 + 8)}^{2024}$

$= {}^{2024}{C_{0}} {(420)}^{2024} + {}^{2024}{C_{1}} {(420)}^{2023} {(8)}^{1} + \dots + {}^{2024}{C_{2023}} (420) {(8)}^{2023} + {}^{2024}{C_{2024}} {(8)}^{2024}$

$= 420$ [Some integer] $+ 64^{1012}$

$= 21 \times 20$ [Some integer] $+ {(63 + 1)}^{1012}$

$= 21 \times 20$ [Some integer] $+ 63 \times$ Some integer + 1

$= 21 \times 20$ [Some integer] $+ 21 \times 3 \times$ Some integer + 1

Hence, remainder = 1

Q 23 :

If the coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}; x \neq 0$ is $α,$ then $| α |$ equals ______ . [2024]

(678)

We have, ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}, x \neq 0$

$= (1 + \frac{6}{x} + \frac{15}{x^{2}} + \frac{20}{x^{3}} + \frac{15}{x^{4}} + \frac{6}{x^{5}} + \frac{1}{x^{6}}) {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}$

$= (1 + \frac{6}{x} + \frac{15}{x^{2}} + \frac{20}{x^{3}} + \frac{15}{x^{4}} + \frac{6}{x^{5}} + \frac{1}{x^{6}}) \times (1 + 7 x^{2} + 21 x^{4} + 35 x^{6} + 35 x^{8} + 21 x^{10} + 7 x^{12} + x^{14})$

$1 - 8 x^{3} + 28 x^{6} - 56 x^{9} + 70 x^{12} - 56 x^{15} + 28 x^{18} - 8 x^{21} + x^{24}$

Coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}$

$= 28 \times 7 + 28 \times 15 - 8 \times 21 \times 6 - 8 \times 20 \times 7 - 8 \times 6 + 15 \times 35 + 35 + 15 \times 21 + 7$

$= 196 + 420 - 1008 - 1120 - 48 + 525 + 35 + 315 + 7$

$= - 678$

$∴$ $| α | = 678$

Q 24 :

If $\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} = \frac{n}{m}$ with $\gcd (n, m) = 1,$ then $n + m$ is equal to _______ . [2024]

(2041)

$\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10}$

$= ({}^{11}{C_{0}} + \frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} + \frac{{}^{11}{C_{10}}}{11} + \frac{{}^{11}{C_{11}}}{12})$ $- 1 - 1 - \frac{1}{12}$

$= \frac{2^{11 + 1} - 1}{12} - 2 - \frac{1}{12} = \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6}$

$∴$ $n + m = 2035 + 6 = 2041$

Q 25 :

Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

(1)

We have, $({{(64)}^{32})}^{32} = ({{(8^{2})}^{32})}^{32} = 8^{2048}$

$= {(9 - 1)}^{2048}$

$= 9^{2048} - {}^{2048}{C_{1}} 9^{2047} + \dots + {}^{2048}{C_{2048} = 9 k + 1}$

So, remainder = 1

Q 26 :

Let $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2}) .$ If $5 α = 6 β,$ then $n$ equals ________ . [2024]

(10)

We have, $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2})$

Now, $α = \sum_{k = 0}^{n} \frac{{}^{n}{C_{k}}}{k + 1} \cdot {}^{n}{C_{k}}$

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

$= \frac{1}{n + 1} \cdot {}^{2 n + 1}{C_{n + 1}}$

Now, $β = \sum_{k = 0}^{n - 1} {}^{n}{C_{n - k} \cdot} \frac{n + 1}{(k + 2) (n + 1)} \cdot {}^{n}{C_{k + 1}}$

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

$∴$ $\frac{β}{α} = \frac{{}^{2 n + 1}{C_{n + 2}}}{{}^{2 n + 1}{C_{n + 1}}} = \frac{(2 n + 1 - n - 1)! \cdot (n + 1)!}{(2 n + 1 - n - 2)! \cdot (n + 2)!}$

$= \frac{n! \cdot (n + 1)!}{(n - 1)! (n + 2) (n + 1)!} = \frac{n}{n + 2} = \frac{5}{6}$

$\Rightarrow n = 10$

Q 27 :

If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

11
15
20
24

1

2

3

4

(3)

Consider $\frac{10^{r + 1} - 1}{10^{r}} = \frac{10^{r + 1}}{10^{r}} - \frac{1}{10^{r}} = 10 - \frac{1}{10^{r}}$

$∴ \sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) {}^{11}C_{r + 1} = \sum_{r = 0}^{10} (10 - \frac{1}{10^{r}}) {}^{11}C_{r + 1}$

$= \sum_{r = 0}^{10} 10 {}^{11}C_{r + 1} - \sum_{r = 0}^{10} \frac{1}{10^{r}} {}^{11}C_{r + 1}$

Now, $\sum_{r = 0}^{10} 10 {}^{11}C_{r + 1} = 10 \sum_{r = 0}^{10} {}^{11}C_{r + 1} + {}^{11}C_{0} - {}^{11}C_{0}$

$= 10 (2^{11} - 1)$ ... (i)

Also, $\sum_{r = 0}^{10} \frac{1}{10^{r}} {}^{11}C_{r + 1} = \sum_{k = 1}^{11} \frac{1}{10^{k - 1}} {}^{11}C_{k} = 10 \sum_{k = 1}^{11} \frac{1}{10^{k}} {}^{11}C_{k}$ ... (ii)

Consider ${(1 + \frac{1}{10})}^{11} = \sum_{k = 0}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k}$

$= {}^{11}C_{0} + \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k} = 1 + \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k}$

$\Rightarrow \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k} = {(\frac{11}{10})}^{11} - 1$

Substituting in (ii), we get

$\sum_{r = 0}^{10} {}^{11}C_{r + 1} {(\frac{1}{10})}^{r} = 10 ({(\frac{11}{10})}^{11} - 1)$ ... (iii)

On combining (i) and (iii), we get

$\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) {}^{11}C_{r + 1} = 10 (2^{11} - 1) - 10 ({(\frac{11}{10})}^{11} - 1)$

$= 10 \times 2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10$

$= 10 \times 2^{11} - \frac{11^{11}}{10^{10}}$

$= \frac{{(20)}^{11} - 11^{11}}{10^{10}} \Rightarrow α = 20$ .

Q 28 :

If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

27
18
81
9

1

2

3

4

(3)

We have, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$

Now, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{r}{2^{r}}) \cdot {}^{9}C_{r} + \sum_{r = 1}^{9} (\frac{3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{9}{2^{r}}) \cdot {}^{8}C_{r - 1} + 3 \sum_{r = 1}^{9} {}^{9}C_{r} {(\frac{1}{2})}^{r}$

$= \frac{9}{2} \sum_{r = 1}^{9} {}^{8}C_{r - 1} {(\frac{1}{2})}^{r - 1} + 3 (\sum_{r = 0}^{9} ({}^{9}C_{r} {(\frac{1}{2})}^{r}) - 1)$

$= \frac{9}{2} {(1 + \frac{1}{2})}^{8} + 3 ({(1 + \frac{1}{2})}^{9} - 1)$

$= \frac{9}{2} \cdot {(\frac{3}{2})}^{8} + 3 {(\frac{3}{2})}^{9} - 3$

$= 6 \cdot {(\frac{3}{2})}^{9} - 3$

On comparing, we get

$\Rightarrow α = 6 and β = 3$

$∴ {(α + β)}^{2} = 81$ .

Q 29 :

The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

3763
18817
16923
33845

1

2

3

4

(2)

We have,

${(2 + \sqrt{3})}^{8} = {}^{8}C_{0} 2^{8} + {}^{8}C_{1} 2^{7} (\sqrt{3}) + . . . + {}^{8}C_{8} {(\sqrt{3})}^{8}$

But, for rational terms, we take only those terms whose exponent is an even number.

Required sum = ${}^{8}C_{0} {(2)}^{8} + {}^{8}C_{2} {(2)}^{6} \cdot {(\sqrt{3})}^{2} + {}^{8}C_{4} {(2)}^{4} {(\sqrt{3})}^{4} + {}^{8}C_{6} {(2)}^{2} {(\sqrt{3})}^{6} + {}^{8}C_{8} {(\sqrt{3})}^{8}$

= $2^{8} + 28 \times 2^{6} \times 3 + 70 \times 2^{4} \times 9 + 28 \times 2^{2} \times 27 + 81$

= 256 + 5376 + 10080 + 3024 + 81 = 18817.

Q 30 :

If $1^{2} \cdot ({}^{15}C_{1}) + 2^{2} \cdot ({}^{15}C_{2}) + 3^{2} \cdot ({}^{15}C_{3}) + . . . + 15^{2} \cdot ({}^{15}C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where m, n, k $\in$ N, then m + n + k is equal to : [2025]

21
18
20
19

1

2

3

4

(4)

Using formula, $\sum_{r = 1}^{n} r^{2} {}^{n}C_{r} = n (n + 1) 2^{n - 2}$

For n = 15

$\sum_{r = 1}^{15} r^{2} {}^{15}C_{r} = 15 \times 16 \times 2^{13}$

= $3 \times 5 \times 2^{4} \times 2^{13}$

= $2^{17} \times 3^{1} \times 5^{1}$

On comparing terms, we get m = 17, n = 1 and k = 1

Thus, m + n + k = 19.

Topic Question Set

Q 1 :

Among the statements

$(S 1) : 2023^{2022} - 1999^{2022} i s d i v i s i b l e b y 8$

$(S 2) : 13 {(13)}^{n} - 11 n - 13 i s d i v i s i b l e b y 144 f o r i n f i n i t e l y m a n y n \in ℕ$ [2023]

Q 2 :

$25^{190} - 19^{190} - 8^{190} + 2^{190}$ is divisible by [2023]

Q 3 :

Let the number ${(22)}^{2022} + {(2022)}^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ is equal to [2023]

Q 4 :

If $\frac{1}{n + 1} {}^{n}{C_{n}} + \frac{1}{n} {}^{n}{C_{n - 1}} + . . . . + \frac{1}{2} {}^{n}{C_{1}} + {}^{n}{C_{0}} = \frac{1023}{10}, t h e n n i s e q u a l t o$ [2023]

Q 5 :

The sum of the coefficients of the first 50 terms in the binomial expansion of ${(1 - x)}^{100},$ is equal to [2023]

Q 6 :

Fractional part of the number $\frac{4^{2022}}{15}$ is equal to [2023]

Q 7 :

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + . . . + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is [2023]

Q 8 :

The value of $\sum_{r = 0}^{22} {}^{22}{C_{r} {}^{23}{C_{r}}}$ is [2023]

Q 9 :

If ${({}^{30}{C_{1}})}^{2} + 2 {({}^{30}{C_{2}})}^{2} + 3 {({}^{30}{C_{3}})}^{2} + . . . + 30 {({}^{30}{C_{30}})}^{2} = \frac{α 60!}{{(30!)}^{2}},$ then $α$ is equal to [2023]

Q 10 :

Let $x = {(8 \sqrt{3} + 13)}^{13}$ and $y = {(7 \sqrt{2} + 9)}^{9} .$ If $[t]$ denotes the greatest integer $\leq t$ , then [2023]

Q 11 :

The largest natural number n such that $3^{n}$ divides 66! is _______ . [2023]

Q 12 :

The remainder, when $7^{103}$ is divided by 17, is ________ . [2023]

Q 13 :

The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

Q 14 :

Suppose $\sum_{r = 0}^{2023} r^{2} {}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ___________ . [2023]

Q 15 :

Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

Q 16 :

The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

Q 17 :

50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

Q 18 :

The remainder on dividing $5^{99}$ by 11 is _________ . [2023]

Q 19 :

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of ${(1 + x)}^{n}$ are in the arithmetic progression, then the maximum value of $n$ is: [2024]

Q 20 :

The coefficient of $x^{70}$ in $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + x^{4} {(1 + x)}^{96} + \dots + x^{54} {(1 + x)}^{46}$ is ${}^{99}{C_{p} - {}^{46}{C_{q} .}}$ Then a possible value of $p + q$ is: [2024]

Q 21 :

Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

Q 22 :

The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

Q 23 :

If the coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}; x \neq 0$ is $α,$ then $| α |$ equals ______ . [2024]

Q 24 :

If $\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} = \frac{n}{m}$ with $\gcd (n, m) = 1,$ then $n + m$ is equal to _______ . [2024]

Q 25 :

Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

Q 26 :

Let $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2}) .$ If $5 α = 6 β,$ then $n$ equals ________ . [2024]

Q 27 :

If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

Q 28 :

If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

Q 29 :

The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

Q 30 :

If $1^{2} \cdot ({}^{15}C_{1}) + 2^{2} \cdot ({}^{15}C_{2}) + 3^{2} \cdot ({}^{15}C_{3}) + . . . + 15^{2} \cdot ({}^{15}C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where m, n, k $\in$ N, then m + n + k is equal to : [2025]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Among the statements (S1):20232022-19992022 is divisible by 8 (S2):13(13)n-11n-13 is divisible by 144 for infinitely many n∈ℕ [2023]

Q 2 : 25190-19190-8190+2190 is divisible by [2023]

Q 3 : Let the number (22)2022+(2022)22 leave the remainder α when divided by 3 and β when divided by 7. Then (α2+β2) is equal to [2023]

Q 4 : If 1n+1Cnn+1nCn-1n+....+12C1n+C0n=102310, then n is equal to [2023]

Q 5 : The sum of the coefficients of the first 50 terms in the binomial expansion of (1-x)100, is equal to [2023]

Q 6 : Fractional part of the number 4202215 is equal to [2023]

Q 7 : The value of 11! 50!+13! 48!+15! 46!+...+149! 2!+151! 1! is [2023]

Q 8 : The value of ∑r=022Cr Cr2322 is [2023]

Q 9 : If (C130)2+2(C230)2+3(C330)2+...+30(C3030)2=α60!(30!)2, then α is equal to [2023]

Q 10 : Let x=(83+13)13 and y=(72+9)9. If [t] denotes the greatest integer ≤t, then [2023]

Q 11 : The largest natural number n such that 3n divides 66! is _______ . [2023]

Q 12 : The remainder, when 7103 is divided by 17, is ________ . [2023]

Q 13 : The remainder, when 19200+23200 is divided by 49, is ________ . [2023]

Q 14 : Suppose ∑r=02023 r2 Cr2023=2023×α×22022. Then the value of α is ___________ . [2023]

Q 15 : Let the sum of the coefficients of the first three terms in the expansion of (x-3x2)n, x≠0,n∈N, be 376. Then the coefficient of x4 is ___________ . [2023]

Q 16 : The remainder when (2023)2023 is divided by 35 is __________ . [2023]

Q 17 : 50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

Q 18 : The remainder on dividing 599 by 11 is _________ . [2023]

Q 19 : If the coefficients of x4,x5 and x6 in the expansion of (1+x)n are in the arithmetic progression, then the maximum value of n is: [2024]

Q 20 : The coefficient of x70 in x2(1+x)98+x3(1+x)97+x4(1+x)96+…+x54(1+x)46 is Cp-Cq.4699 Then a possible value of p+q is: [2024]

Q 21 : Let α=∑r=0n(4r2+2r+1)Crn and β=(∑r=0nCrnr+1)+1n+1. If 140<2αβ<281, then the value of n is ________. [2024]

Q 22 : The remainder when 4282024 is divided by 21 is ______ . [2024]

Q 23 : If the coefficient of x30 in the expansion of (1+1x)6(1+x2)7(1-x3)8; x≠0 is α, then |α| equals ______ . [2024]

Q 24 : If C1112+C2113+⋯+C91110=nm with gcd(n,m)=1, then n+m is equal to _______ . [2024]

Q 25 : Remainder when 643232 is divided by 9 is equal to _____ . [2024]

Q 26 : Let α=∑k=0n((Ckn)2k+1) and β=∑k=0n-1(Ck Ck+1nnk+2). If 5α=6β, then n equals ________ . [2024]

Q 27 : If ∑r=010(10r+1–110r)·Cr+111=α11–11111010, then α is equal to : [2025]

Q 28 : If ∑r=19(r+32r)·Cr9=α(32)9–β, α,β∈ℕ, then (α+β)2 is equal to [2025]

Q 29 : The sum of all rational terms in the expansion of (2+3)8 is equal to [2025]

Q 30 : If 12·(C115)+22·(C215)+32·(C315)+...+152·(C1515)=2m·3n·5k, where m, n, k ∈ N, then m + n + k is equal to : [2025]

Want Us to Email you About Special Offers & Updates?

Q 1 :

Among the statements

$(S 1) : 2023^{2022} - 1999^{2022} i s d i v i s i b l e b y 8$

$(S 2) : 13 {(13)}^{n} - 11 n - 13 i s d i v i s i b l e b y 144 f o r i n f i n i t e l y m a n y n \in ℕ$ [2023]

Q 2 :

$25^{190} - 19^{190} - 8^{190} + 2^{190}$ is divisible by [2023]

Q 3 :

Let the number ${(22)}^{2022} + {(2022)}^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ is equal to [2023]

Q 4 :

If $\frac{1}{n + 1} {}^{n}{C_{n}} + \frac{1}{n} {}^{n}{C_{n - 1}} + . . . . + \frac{1}{2} {}^{n}{C_{1}} + {}^{n}{C_{0}} = \frac{1023}{10}, t h e n n i s e q u a l t o$ [2023]

Q 5 :

The sum of the coefficients of the first 50 terms in the binomial expansion of ${(1 - x)}^{100},$ is equal to [2023]

Q 6 :

Fractional part of the number $\frac{4^{2022}}{15}$ is equal to [2023]

Q 7 :

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + . . . + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is [2023]

Q 8 :

The value of $\sum_{r = 0}^{22} {}^{22}{C_{r} {}^{23}{C_{r}}}$ is [2023]

Q 9 :

If ${({}^{30}{C_{1}})}^{2} + 2 {({}^{30}{C_{2}})}^{2} + 3 {({}^{30}{C_{3}})}^{2} + . . . + 30 {({}^{30}{C_{30}})}^{2} = \frac{α 60!}{{(30!)}^{2}},$ then $α$ is equal to [2023]

Q 10 :

Let $x = {(8 \sqrt{3} + 13)}^{13}$ and $y = {(7 \sqrt{2} + 9)}^{9} .$ If $[t]$ denotes the greatest integer $\leq t$ , then [2023]

Q 11 :

The largest natural number n such that $3^{n}$ divides 66! is _______ . [2023]

Q 12 :

The remainder, when $7^{103}$ is divided by 17, is ________ . [2023]

Q 13 :

The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

Q 14 :

Suppose $\sum_{r = 0}^{2023} r^{2} {}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ___________ . [2023]

Q 15 :

Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

Q 16 :

The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

Q 17 :

50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

Q 18 :

The remainder on dividing $5^{99}$ by 11 is _________ . [2023]

Q 19 :

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of ${(1 + x)}^{n}$ are in the arithmetic progression, then the maximum value of $n$ is: [2024]

Q 20 :

The coefficient of $x^{70}$ in $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + x^{4} {(1 + x)}^{96} + \dots + x^{54} {(1 + x)}^{46}$ is ${}^{99}{C_{p} - {}^{46}{C_{q} .}}$ Then a possible value of $p + q$ is: [2024]

Q 21 :

Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

Q 22 :

The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

Q 23 :

If the coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}; x \neq 0$ is $α,$ then $| α |$ equals ______ . [2024]

Q 24 :

If $\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} = \frac{n}{m}$ with $\gcd (n, m) = 1,$ then $n + m$ is equal to _______ . [2024]

Q 25 :

Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

Q 26 :

Let $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2}) .$ If $5 α = 6 β,$ then $n$ equals ________ . [2024]

Q 27 :

If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

Q 28 :

If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

Q 29 :

The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

Q 30 :

If $1^{2} \cdot ({}^{15}C_{1}) + 2^{2} \cdot ({}^{15}C_{2}) + 3^{2} \cdot ({}^{15}C_{3}) + . . . + 15^{2} \cdot ({}^{15}C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where m, n, k $\in$ N, then m + n + k is equal to : [2025]