If \(26\left(\frac{2^{3}}{3}\left({ }^{12} \mathrm{C}_{2}\right)+\frac{2^{5}}{5}\left({ }^{12} \mathrm{C}_{4}\right)+\frac{2^{7}}{7}\left({ }^{12} \mathrm{C}_{6}\right)+\ldots .+\frac{2^{13}}{13}\left({ }^{12} \mathrm{C}_{12}\right)\right) =3^{13}-\alpha\), then \(\alpha\) is equal to :

Binomial Theorem — JEE Main · 2026 · 08 Apr 2026 · Shift 2 Question with Solution

26 (\frac{2 ^{3}}{3} (^{12} C_{2}) + \frac{2 ^{5}}{5} (^{12} C_{4}) + \frac{2 ^{7}}{7} (^{12} C_{6}) + \dots . + \frac{2 ^{13}}{13} (^{12} C_{12})) = 3^{13} - α

, then

α

is equal to :

Answer: (C)

Expand

(1 + x)^{12} = \sum_{k = 0}^{12} (k 12) x^{k}

and integrate from

- 2

2

. Odd powers integrate to zero, while each even power gives

\int_{- 2}^{2} x^{k} d x = \frac{2 ^{k + 2}}{k + 1}

, so

\int_{- 2}^{2} (1 + x)^{12} d x = 2 j = 0 \sum 6 (2 j 12) \frac{2 ^{2 j + 1}}{2 j + 1} .

The left side equals

[\frac{( 1 + x ) ^{13}}{13}]_{- 2}^{2} = \frac{3 ^{13} + 1}{13}

. Writing the right side out,

\frac{3 ^{13} + 1}{13} = 2 (2 + \frac{2 ^{3}}{3} (2 12) + \frac{2 ^{5}}{5} (4 12) + \dots + \frac{2 ^{13}}{13} (12 12)),

3^{13} + 1 = 52 + 26 (\frac{2 ^{3}}{3} (2 12) + \dots + \frac{2 ^{13}}{13} (12 12))

. Rearranging,

26 (\frac{2 ^{3}}{3} (2 12) + \frac{2 ^{5}}{5} (4 12) + \dots + \frac{2 ^{13}}{13} (12 12)) = 3^{13} - 51,

hence

α = 51.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option