JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
If $26\bigg(\frac{2^{3}}{3}\left({}^{12}C_{2}\right)+\frac{2^{5}}{5}\left({}^{12}C_{4}\right)+$ $\frac{2^{7}}{7}\left({}^{12}C_{6}\right)+\ldots+\frac{2^{13}}{13}\left({}^{12}C_{12}\right)\bigg)$ $=3^{13}-\alpha$, then $\alpha$ is equal to :
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. Turn $\frac{1}{r+1}\,{}^{12}C_r$ into a coefficient of $(1+x)^{13}$; then the bracket is half of a known sum. Formulas used. (1) Absorption: $\dfrac{{}^{n}C_r}{r+1}=\dfrac{{}^{n+1}C_{r+1}}{n+1}$(2) Odd-index sum: $\displaystyle\sum_{k\ \text{odd}}{}^{n}C_k\,x^k=\frac{(1+x)^n-(1-x)^n}{2}$Step 1 - General term. The terms have even $r=2,4,\dots,12$ and look like $\dfrac{2^{r+1}}{r+1}\,{}^{12}C_r$. By (1), $$\frac{2^{r+1}}{r+1}\,{}^{12}C_r=\frac{2^{r+1}\,{}^{13}C_{r+1}}{13}.$$ Put $k=r+1$: $k$ runs over the odd numbers $3,5,\dots,13$. Step 2 - Odd-index sum. By (2) with $n=13$, $x=2$: $$\sum_{k\ \text{odd}}{}^{13}C_k\,2^k=\frac{3^{13}-(-1)^{13}}{2}=\frac{3^{13}+1}{2}.$$ Our sum starts at $k=3$, so remove the $k=1$ term ${}^{13}C_1\cdot2=26$: $$\text{bracket}=\frac{1}{13}\left(\frac{3^{13}+1}{2}-26\right).$$ Step 3 - Multiply by $26$. $$26\times\text{bracket}=2\left(\frac{3^{13}+1}{2}-26\right)=3^{13}+1-52=3^{13}-51.$$ So $\alpha=51$. Answer: (C)