JEE Main 2025 Jan 22 Shift 1, Mathematics Q22: Properties Of Binomial Coefficient

Use $\frac{\left(\genfrac{}{}{0ex}{}{11}{k}\right)}{k+1}=\frac{1}{12}\left(\genfrac{}{}{0ex}{}{12}{k+1}\right)$, valid since $\left(\genfrac{}{}{0ex}{}{12}{k+1}\right)=\frac{12}{k+1}\left(\genfrac{}{}{0ex}{}{11}{k}\right)$.

With $k=2r+1$ (odd values $1,3,5,7,9,11$), the sum becomes

$\sum _{r=0}^5\frac{\left(\genfrac{}{}{0ex}{}{11}{2r+1}\right)}{2r+2}=\frac{1}{12}\sum _{r=0}^5\left(\genfrac{}{}{0ex}{}{12}{2r+2}\right)=\frac{1}{12}\left[\left(\genfrac{}{}{0ex}{}{12}{2}\right)+\left(\genfrac{}{}{0ex}{}{12}{4}\right)+\left(\genfrac{}{}{0ex}{}{12}{6}\right)+\left(\genfrac{}{}{0ex}{}{12}{8}\right)+\left(\genfrac{}{}{0ex}{}{12}{10}\right)+\left(\genfrac{}{}{0ex}{}{12}{12}\right)\right]$.

This is the sum of even-index binomials from $\left(\genfrac{}{}{0ex}{}{12}{2}\right)$ to $\left(\genfrac{}{}{0ex}{}{12}{12}\right)$. The sum of all even-index $\left(\genfrac{}{}{0ex}{}{12}{k}\right)$ is $2^{11}=2048$; subtracting $\left(\genfrac{}{}{0ex}{}{12}{0}\right)=1$ gives $2047$.

Sum $=\frac{2047}{12}$. Since $\gcd (2047,12)=1$, we get $m=2047$, $n=12$, and $m-n=2035$.