Suppose A and B are the coefficients of and terms respectively in the binomial expansion of . If 2A = 5B, then n is equal to: [2025]

Question

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]

Smart Achievers · Accepted Answer

(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.

Solution

Q.
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]

1 22

2 19

3 21

4 20

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