JEE Main 2025 — Binomial Theorem Question with Solution

$\begin{array}{rl}& (1+x{)}^{11}={}^{11}{C}_0+{}^{11}{C}_{1}x+{}^{11}{C}_2{x}^2+\cdots +{}^{11}{C}_{11}{x}^{11}\\ & {\int }_0^1(1+x{)}^{11}dx={\int }_0^1\left({}^{11}{C}_0+{}^{11}{C}_{1}x+{}^{11}{C}_{2}x{}^2+\cdots +{}^{11}{C}_{11}x{}^{11}\right)dx\\ & {{\frac{(1-x{)}^{12}}{12}]}_0^1={}^{11}{C}_{0x}+\frac{{}^{11}{C}_1{x}^2}{2}+\frac{{}^{11}{C}_{2}x}{3}+\cdots +\frac{{}^{11}{C}_{11}{x}^{12}}{12}]}_0^1\end{array}$

$$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots +\frac{C_{11}}{12}\cdots$$ ........(1)

Now,

$$\begin{array}{rl}& {\int }_{-1}^0(1+x{)}^{11}dx={\int }_{-1}^0\left({}^{11}{C}_0+{}^{11}{C}_{1}x+{}^{11}{C}_2{x}^2+\cdots +{}^{11}{C}_{11}{x}^{11}\right)dx\\ & {{\frac{(1+x{)}^{12}}{12}]}_{-1}^0={}^{11}{C}_{0}x+\frac{{}^{11}{C}_1{x}^2}{2}+\frac{{}^{11}{C}_2{x}^3}{3}+\cdots +\frac{{}^{11}{C}_{11}{x}^{12}}{12}]}_{-1}^0\end{array}$$

$$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3}\cdots$$ …….(2)

$\begin{array}{rl}& \text{ (1) }-(2)\\ & \\ & =\frac{2^{12}-2}{12}=2\left[\frac{C_1}{2}+\frac{C_3}{4}+\cdots \right]\\ & \\ & \Rightarrow \sum _{r=0}^5\frac{C_{2r+1}}{2r+2}=\frac{2^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n}\\ & \\ & =2047-12=2035\end{array}$