JEE Main 2024 — 30 January, Shift 1 — Question 26

$$\left(7^{\frac12}+11^{\frac16}\right)^{824}$$

The general term in the binomial expansion is

$$T_{r+1}=\binom{824}{r}\left(7^{\frac12}\right)^{824-r}\left(11^{\frac16}\right)^r = \binom{824}{r}\,7^{\frac{824-r}{2}}\,11^{\frac{r}{6}}.$$

For the term to be integral, the exponents of both $7$ and $11$ must be integers.

$\frac{824-r}{2}\in \mathbb{Z}$ $\Rightarrow$ $r$ is even

$\frac{r}{6}\in \mathbb{Z}$ $\Rightarrow$ $r$ is a multiple of $6$

Hence,

$$r=0,6,12,18,\ldots,822$$

The number of such values of $r$ is

$$\frac{822}{6}+1=137+1=138.$$ $$\boxed{138}$$