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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Binomial Theorem

Suppose A and B are the coefficients of 30^ th and 12^ th terms...

Suppose A and B are the coefficients of 30th 30^{\text{th }} and 12th 12^{\text{th }} terms respectively in the binomial expansion of (1+x)2n−1(1+x)^{2 \mathrm{n}-1}. If 2A=5B,thennisequalto:2 \mathrm{ A}=5 \mathrm{ B} , then n is equal to :

Options

  1. A22
  2. B20
  3. C21
  4. D19

Correct answer

C. 21

Step-by-step solution

A=2n−1C29 B=2n−1C1122n−1C29=52n−1C112(2n−1)!29!(2n−30)!=5(2n−1)!(2n−12)!11!129…12⋅5=1(2n−12)(2n−13)…(2n−29)2130⋅29…12=1(2n−12)(2n−13)…(2n−29)122n−12=30n=21\begin{aligned} & \mathrm{A}={ }^{2 \mathrm{n}-1} \mathrm{C}_{29} \quad \mathrm{~B}={ }^{2 \mathrm{n}-1} \mathrm{C}_{11} \\ & 2^{2 \mathrm{n}-1} \mathrm{C}_{29}=5{ }^{2 \mathrm{n}-1} \mathrm{C}_{11} \\ & 2 \frac{(2 \mathrm{n}-1)!}{29!(2 \mathrm{n}-30)!}=5 \frac{(2 \mathrm{n}-1)!}{(2 \mathrm{n}-12)!11!} \\ & \frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 \mathrm{n}-12)(2 \mathrm{n}-13) \ldots(2 \mathrm{n}-29) 2} \\ & \frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 \mathrm{n}-12)(2 \mathrm{n}-13) \ldots(2 \mathrm{n}-29) 12} \\ & 2 \mathrm{n}-12=30 \\ & \mathrm{n}=21\end{aligned}
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