JEE Mathematics 2025 Question with Solution

Given: In the expansion of $(1+x)^{2n-1}$, A is the coefficient of the 30th term and B is the coefficient of the 12th term. Also, $2A=5B$.

Find: The value of $n$.

For the binomial expansion of $(1+x)^{2n-1}$, the general term is

$$T_{k+1}=\binom{2n-1}{k}x^k$$

So, the 30th term corresponds to $k=29$ and the 12th term corresponds to $k=11$.

Hence,

$$A=\binom{2n-1}{29}, \qquad B=\binom{2n-1}{11}$$

Using the given relation,

$$2\binom{2n-1}{29}=5\binom{2n-1}{11}$$

Thus,

$$\frac{\binom{2n-1}{29}}{\binom{2n-1}{11}}=\frac{5}{2}$$

the solution concludes that solving this relation gives

$$n=21$$

Therefore, the correct option is C.

Given: The expansion is $(1+x)^{2n-1}$ and $2A=5B$, where A and B are coefficients of the 30th and 12th terms respectively.

Find: The value of $n$.

The general term is

$$T_{k+1}=\binom{2n-1}{k}x^k$$

So,

1. For the 30th term, $k=29$. Therefore,

$$A=\binom{2n-1}{29}$$

2. For the 12th term, $k=11$. Therefore,

$$B=\binom{2n-1}{11}$$

Now use

$$2A=5B$$

Substituting the coefficients,

$$2\binom{2n-1}{29}=5\binom{2n-1}{11}$$

So,

$$\frac{\binom{2n-1}{29}}{\binom{2n-1}{11}}=\frac{5}{2}$$

Using the factorial form mentioned in the solution,

$$\frac{\binom{m}{r}}{\binom{m}{s}}=\frac{s!(m-s)!}{r!(m-r)!}$$

with $m=2n-1$, $r=29$ and $s=11$.

The provided solution states that after simplification, the integer solution is

$$21$$

Therefore, $n=21$ and the correct option is C.