Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+{x}^{n}),\text{ }n\in N,\text{ }0\leq r\leq n.$ If ${P}_{n}={C}_{0}−{C}_{1}+\frac{{2}^{2}}{3}{C}_{2}−\frac{{2}^{3}}{4}{C}_{3}+...+\frac{{(−2)}^{n}}{n+1}{C}_{n},$ then the value of $\sum _{n=1}^{25}\frac{1}{{P}_{2n}}$ equals [JEE Main 2026, 22 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: a) $675$. ${P}_{n}={\sum }_{r=0}^{n}\frac{{}^{n}{C}_{r}{(−2)}^{r}}{r+1}$ $={\sum }_{r=0}^{n}\frac{1}{\left(n+1\right)}{}^{n+1}{C}_{r+1}{(−2)}^{r}$ $=\frac{−1}{2\left(n+1\right)}{\sum }_{r=0}^{n}{}^{n+1}{C}_{r+1}{(−2)}^{r+1}$ $=\frac{−1}{2\left(\text{n}+1\right)}\left[{(1−2)}^{\text{n}+1}−1\right]$ ${\text{⇒P}}_{\text{n}}=\frac{1}{2\left(\text{n}+1\right)}\left[1−{(−1)}^{\text{n}+1}\right]$ ${P}_{2n}=\frac{1}{2\left(2n+1\right)}\left[1−{(−1)}^{2n+1}\right]$ ${P}_{2n}=\frac{1}{2n+1}$ $\sum_{n=1}^{25} \frac{1}{P_{2 n}}=\sum_{n=1}^{25}(2 n+1)=3+5+\ldots . .+51=\frac{25}{2}[51+3]=25 \times 27=675$

Let \(C_r\) denote the coefficient of \(x^r\) in the binomial expansion of \((1+{x}^{n}),\text{ }n\in N,\text{ }0\leq r\…

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