JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Binomial Theorem

The general term is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{1016}{r}\right)5^{\frac{1016-r}{2}}{7}^{\frac{r}{8}}$. For an integral term both exponents must be integers. The exponent $\frac{r}{8}$ is an integer only when $r$ is a multiple of $8$; for such $r$, $1016-r$ is even, so $\frac{1016-r}{2}$ is automatically an integer. Thus $r=0,8,16,\dots ,1016$, an arithmetic progression with $\frac{1016-0}{8}+1=128$ terms. The number of integral terms is $128$.