JEE Main 2026 (Online) 22nd January Evening ShiftQuestion 1
Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$, $n\in\mathbf{N}$, $0\le r\le n$. 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 $\sum_{n=1}^{25}\frac{1}{P_{2n}}$ equals :
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. A factor $\frac{1}{k+1}$ next to $t^k$ comes from integrating $t^k$ from $0$ to $1$. Formulas used. (1) $\int_0^1 t^k\,dt=\dfrac{1}{k+1}$(2) $\int_0^1(1-2t)^n\,dt=\dfrac{1-(-1)^{n+1}}{2(n+1)}$Step 1 - General term. The $k$-th term is $\dfrac{(-2)^k}{k+1}\,C_k$ (check: $k=1$ gives $-C_1$, $k=2$ gives $\frac43C_2$). By (1), $$P_n=\sum_{k=0}^{n}C_k(-2)^k\int_0^1t^k\,dt=\int_0^1(1-2t)^n\,dt.$$ Step 2 - Integrate. By (2), $$P_n=\frac{1-(-1)^{n+1}}{2(n+1)},\qquad P_{2n}=\frac{1+1}{2(2n+1)}=\frac{1}{2n+1}.$$ Step 3 - Required sum. $$\sum_{n=1}^{25}\frac{1}{P_{2n}}=\sum_{n=1}^{25}(2n+1)=2\cdot\frac{25\cdot26}{2}+25=650+25=675.$$ Answer: (B)