JEE Main 2026 Apr 8 Shift 2, Mathematics Q8: Definite Integral as Limit of a Sum

Start from the binomial expansion $(1+x{)}^{12}={\sum }_{r=0}^{12}{}^{12}{C}_r{x}^r$ and integrate from $0$ to $x$:

$\frac{(1+x{)}^{13}-1}{13}={\sum }_{r=0}^{12}{}^{12}{C}_r\frac{x^{r+1}}{r+1}$

Evaluate at $x=2$ and at $x=-2$ and subtract: the odd-$r$ terms cancel and the even-$r$ terms double.

At $x=2$: $\frac{3^{13}-1}{13}$.

At $x=-2$: $\frac{(-1{)}^{13}-1}{13}=\frac{-2}{13}$.

Difference: $\frac{3^{13}-1}{13}-\frac{-2}{13}=\frac{3^{13}+1}{13}=2{\sum }_{r\text{ even}}{}^{12}{C}_r\frac{2^{r+1}}{r+1}$.

The $r=0$ term equals $2\cdot 1\cdot 2=4$; pulling it out and isolating the sum starting from $r=2$:

$\frac{3^{13}+1}{13}-4=2{\sum }_{r=2,4,\dots ,12}{}^{12}{C}_r\frac{2^{r+1}}{r+1}$

Multiplying both sides by $13$ and rearranging gives $3^{13}-51=26\left(\frac{2^3}{3}{}^{12}{C}_2+\frac{2^5}{5}{}^{12}{C}_4+\dots \right)$.

$\therefore$ $\alpha =51$.