Let C r denote the coefficient of x r in the binomial expansion of (1+x) n , n in mathbb N , 0 leq r leq n . If P n = C 0 - C 1 +frac 2 2 3 C 2 -frac 2 3 4 C 3 +ldots . .+frac (-2) n n +1 C n , then the value of sum n=1 25 frac 1 P 2 n equals.

Mathematics · Binomial Theorem

JEE Main 2026 — 22 January, Evening Shift — Question 20

Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $\mathrm{x}^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{n}, n \in \mathbb{N}, 0 \leq r \leq n$ .

If $\mathrm{P}_{\mathrm{n}}=\mathrm{C}_{0}-\mathrm{C}_{1}+\frac{2^{2}}{3} \mathrm{C}_{2}-\frac{2^{3}}{4} \mathrm{C}_{3}+\ldots . .+\frac{(-2)^{\mathrm{n}}}{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}}$ , then the value of $\sum_{n=1}^{25} \frac{1}{\mathrm{P}_{2 n}}$ equals.

Option A:
580
Option B:
525
Option C:
650
Option D:
675
Correct

Answer: D

Step-by-step solution

$P_{n}=\sum_{r=0}^{n} \frac{{ }^{n} C_{r}(-2)^{r}}{r+1}=\sum_{r=0}^{n} \frac{1}{(n+1)}{ }^{n+1} C_{r+1}(-2)^{r}$

$=\frac{-1}{2(n+1)} \sum_{r=0}^{n}{ }^{n+1} C_{r+1}(-2)^{r+1}$

$=\frac{-1}{2(\mathrm{n}+1)}\left[(1-2)^{\mathrm{n}+1}-1\right]$

$\mathrm{P}_{\mathrm{n}}=\frac{1}{2(\mathrm{n}+1)}\left[1-(-1)^{\mathrm{n}+1}\right]$

$P_{2 n}=\frac{1}{2(2 n+1)}\left[1-(-1)^{2 n+1}\right]$

$\mathrm{P}_{2 \mathrm{n}}=\frac{1}{2 \mathrm{n}+1}$

$\sum_{n=1}^{25} \frac{1}{P_{2 n}}=\sum_{n=1}^{25}(2 n+1)$

$=3+5+\ldots+51$

$=\frac{25}{2}[51+3]$

$=25 \times 27=675$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients