Q47 / 75 · JEE Main 2025 (22 Jan Shift 1) · Mathematics
If $\sum_{r=0}^5 \frac{{ }^{11} C_{2r+1}}{2 r+2}=\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}-\mathrm{n}$ is equal to _______
Correct answer: 2035
Solution
$\begin{aligned}
& (1+x)^{11}={ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11} \\
& \int_0^1(1+x)^{11} d x=\int_0^1\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x x^{11}\right) d x \\
& \left.\left.\frac{(1-x)^{12}}{12}\right|_0 ^1={ }^{11} C_{0 x}+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_0^1 \\
& \frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{11}}{12} \ldots(1)
\end{aligned}$
Now,
$\begin{aligned}
& \int_{-1}^0(1+x)^{11} d x=\int_{-1}^0\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11}\right) d x \\
& \left.\left.\frac{(1+x)^{12}}{12}\right]_{-1}^0={ }^{11} C_0 x+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x^3}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_{-1}^0
\end{aligned}$
$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3} \cdots$
$\begin{aligned}
& (1)-(2) \\
& =\frac{2^{12}-2}{12}=2\left[\frac{C_1}{2}+\frac{C_3}{4}+\cdots\right] \\
& \Rightarrow \sum_{r=0}^5 \frac{C_{2 r+1}}{2 r+2}=\frac{2^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n} \\
& =2047-12=2035
\end{aligned}$
Hence, the correct answer is 2035.