JEE 2025 — Mathematics PYQ

$$\sum _{r=0}^5\frac{{}^{11}{C}_{2r+1}}{2r+2}$$

$$=\frac{1}{12}\sum _{r=0}^5{}^{12}{C}_{2r+2}\left(\frac{{}^{n+1}{C}_{r+1}}{n+1}=\frac{{}^n{C}_r}{r+1}\right)$$

$$=\frac{1}{12}\left({}^{12}{C}_2+{}^{12}{C}_4+{}^{12}{C}_6+{}^{12}{C}_8+{}^{12}{C}_{10}+{}^{12}{C}_{12}\right)$$

$$=\frac{1}{12}(2^{11}-1)$$

$$=\frac{2^{11}-1}{12}$$

$$=\frac{2047}{12}=\frac{m}{n}$$

$$\Rightarrow m-n=2047-12=2035$$