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

Q 11 :
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 12 :
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.

Q 13 :
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
3
4
8

1

2

3

4

(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 14 :
The remainder when $({{(64)}^{(64)})}^{(64)}$ is divided by 7 is equal to [2025]

3
1
6
4

1

2

3

4

(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 15 :
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
8
20
13

1

2

3

4

(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 16 :
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
19
21
20

1

2

3

4

(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 17 :
The remainder, when $7^{103}$ is divided by 23, is equal to: [2025]

6
14
9
17

1

2

3

4

(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 18 :
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]

0%

(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 19 :
The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]

0%

(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 = $7 \times 9 = 63$ .

Topic Question Set

Q 11 :
The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

Q 12 :
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]

Q 13 :
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 14 :
The remainder when $({{(64)}^{(64)})}^{(64)}$ is divided by 7 is equal to [2025]

Q 15 :
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 16 :
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 17 :
The remainder, when $7^{103}$ is divided by 23, is equal to: [2025]

Q 18 :
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]

0%

Q 19 :
The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]

0%

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Topic Question Set

Q 11 : The sum of all rational terms in the expansion of (2+3)8 is equal to [2025]

Q 12 : 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]

Q 13 : 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 14 : The remainder when ((64)(64))(64) is divided by 7 is equal to [2025]

Q 15 : 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 16 : 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 17 : The remainder, when 7103 is divided by 23, is equal to: [2025]

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

0%

Q 19 : The product of the last two digits of (1919)1919 is __________. [2025]

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Q 11 :
The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

Q 12 :
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]

Q 13 :
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 14 :
The remainder when $({{(64)}^{(64)})}^{(64)}$ is divided by 7 is equal to [2025]

Q 15 :
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 16 :
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 17 :
The remainder, when $7^{103}$ is divided by 23, is equal to: [2025]

Q 18 :
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 19 :
The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]