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

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(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

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(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

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(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

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(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]

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

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(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]

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

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(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]

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

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(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]

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

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(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
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15
30

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(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

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(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.

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]

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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]

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]