JEE Main 2026 · 8 Apr · Shift 2 · Mathematics Q8

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If $26\left(\dfrac{2^3}3\left({}^{12}C_2\right)+\dfrac{2^5}5\left({}^{12}C_4\right)+\dfrac{2^7}7\left({}^{12}C_6\right)+\dots+\dfrac{2^{13}}{13}\left({}^{12}C_{12}\right)\right)=3^{13}-\alpha$ , then $\alpha$ is equal to :

Answer and worked solution

Official answer: C

Integral form. $\dfrac{2^{k+1}}{k+1}\binom{12}k=\displaystyle\int_0^2\binom{12}kx^k\,dx$ . The even-k terms come from
$\int_0^2\frac{(1+x)^{12}+(1-x)^{12}}2\,dx=\frac{3^{13}-1+1+1}{26}=\frac{3^{13}+1}{26}.$

Remove k = 0. That term is 2, so the bracket is $\dfrac{3^{13}+1}{26}-2$ and
$26(\dots)=3^{13}+1-52=3^{13}-51.$

(C)

Topics:Properties of binomial coefficients Properties of definite integrals

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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 8