MathsJEE Main 8 Apr 2026, Shift 2·Q8·MCQ·Tough

Binomial Theorem · Binomial coefficient sums by integration

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) + \ldots + \frac{2^{13}}{13}\left({}^{12}C_{12}\right)\right) = 3^{13} - \alpha,

then

\alpha

is equal to:

(1)45
(2)48
(3)51
(4)54

We corrected a printing error in the original paper

Source printed the first term as “ $\frac{2^3}{3}({}^{13}C_2)$ ”; corrected to the intended ${}^{12}C_2$ (all other terms are ${}^{12}C_k$ ; with ${}^{13}C_2$ the value is $\alpha = -781$ , no option).

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See the full 8 Apr 2026, Shift 2 paper