JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let $C_{r}$ denote the coefficient of $x^{r}$ in the binomial expansion of $(1 + x)^{n}$ , $n \in N$ , $0 \leq r \leq n$ . If $P_{n} = C_{0} - C_{1} + \frac{2 ^{2}}{3} C_{2} - \frac{2 ^{3}}{4} C_{3} + \dots + \frac{( - 2 ) ^{n}}{n + 1} C_{n}$ , then the value of $n = 1 \sum 25 \frac{1}{P _{2 n}}$ equals.

A
580
B
525
C
650
D
675

Solution

$P_{n} = r = 0 \sum n \frac{^{n} C _{r} ( - 2 ) ^{r}}{r + 1} = \frac{1}{n + 1} r = 0 \sum n^{n + 1} C_{r + 1} (- 2)^{r} = \frac{- 1}{2 ( n + 1 )} r = 0 \sum n^{n + 1} C_{r + 1} (- 2)^{r + 1}$ .

The inner sum equals $(1 - 2)^{n + 1} - 1 = (- 1)^{n + 1} - 1$ , so $P_{n} = \frac{1 - ( - 1 ) ^{n + 1}}{2 ( n + 1 )}$ .

For even index $2 n$ : $(- 1)^{2 n + 1} = - 1$ , hence $P_{2 n} = \frac{1 - ( - 1 )}{2 ( 2 n + 1 )} = \frac{1}{2 n + 1}$ .

Therefore $n = 1 \sum 25 \frac{1}{P _{2 n}} = n = 1 \sum 25 (2 n + 1) = 3 + 5 + \dots + 51 = \frac{25 ( 3 + 51 )}{2} = 25 \cdot 27 = 675$ .

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