If $26\left(\frac{{2}^{3}}{3}\left({}^{12}C_{2}\right)+\frac{{2}^{5}}{5}\left({}^{12}C_{4}\right)+\frac{{2}^{7}}{7}\left({}^{12}C_{6}\right)+⋯+\frac{{2}^{13}}{13}\left({}^{12}C_{12}\right)\right)$$={3}^{13}-\alpha$, then $\alpha$ is equal to: [JEE Main 2026, 8 Apr (Shift 2)]

Question

YESPYQ · Accepted Answer

Correct answer: c) $51$. Let $S=26\left(\frac{2^3}{3}({}^{12}C_2)+\frac{2^5}{5}({}^{12}C_4)+\frac{2^7}{7}({}^{12}C_6)+\cdots+\frac{2^{13}}{13}({}^{12}C_{12})\right)$ $S=26\sum_{r=1}^{6}\frac{2^{2r+1}}{2r+1}({}^{12}C_{2r})$ Using $\frac{{}^{12}C_{2r}}{2r+1}=\frac{{}^{13}C_{2r+1}}{13}$, we get: $S=26\sum_{r=1}^{6}\frac{2^{2r+1}}{13}({}^{13}C_{2r+1})$ $S=2\sum_{r=1}^{6}2^{2r+1}({}^{13}C_{2r+1})$ $S=2\left[\sum_{r=0}^{6}2^{2r+1}({}^{13}C_{2r+1})-2({}^{13}C_1)\right]$ $S=2\sum_{r=0}^{6}2^{2r+1}({}^{13}C_{2r+1})-52$ $\sum_{r=0}^{6}{}^{13}C_{2r+1}2^{2r+1}=\frac{(1+2)^{13}-(1-2)^{13}}{2}$ $S=2\cdot\frac{3^{13}-(-1)^{13}}{2}-52$ $S=3^{13}+1-52=3^{13}-51$ Given $S=3^{13}-\alpha$, so $\alpha=51$

If \(26\left(\frac{{2}^{3}}{3}\left({}^{12}C_{2}\right)+\frac{{2}^{5}}{5}\left({}^{12}C_{4}\right)+\frac{{2}^{7}}{7}\lef…

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