JEE Main 2003 — Binomial Theorem Question with Solution

General term = $${}^{256}{C}_r.{\left(\sqrt{3}\right)}^{256-r}.{\left(\text{root}8\text{of}5\right)}^r$$
= $${}^{256}{C}_r.{\left(3\right)}^{\frac{256-r}{2}}.{\left(5\right)}^{\frac{r}{8}}$$

When $$\frac{256-r}{2}$$ is integer then $${\left(3\right)}^{\frac{256-r}{2}}$$ is integer.
And when $$\frac{r}{8}$$ is integer then $${\left(5\right)}^{\frac{r}{8}}$$ is integer.

Entire general term will be integer when $$\frac{256-r}{2}$$ and $$\frac{r}{8}$$ both are integer.

$$\frac{256-r}{2}$$ is integer when r = 0, 2, 4, 6, ......, 256

$$\frac{r}{8}$$ is integer when r = 0, 8, 16 ,......., 256

Now both $$\frac{256-r}{2}$$ and $$\frac{r}{8}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, t_n = a + (n - 1)d

$$\therefore$$ n = $$\frac{256}{8}+1$$ = 32 + 1 = 33