JEE Main 2020 — Binomial Theorem Question with Solution

General term of the expression,

$$T_{r+1}={}^n{C}_r{\left(3^{\frac{1}{2}}\right)}^{n-r}{\left(5^{\frac{1}{8}}\right)}^r$$

$$={}^n{C}_r{\left(3\right)}^{\frac{n-r}{2}}{\left(5\right)}^{\frac{r}{8}}$$

We will get integral term when $$\frac{n-r}{2}$$ and $$\frac{r}{8}$$ are integer

$$\therefore$$ (1) n $$-$$ r is multiple of 2

$$\Rightarrow$$ n $$-$$ r = 0, 2, 4, ......

(2) r is multiple of 8

$$\Rightarrow$$ r = 0, 8, 16, .......

From this two conditions common values are = 0, 8, 16, ....... which will becomes integral terms.

Given that there are 33 integral terms.

Here first integral term at 0^th position.

Second integral term at 8^th position.

$$\therefore$$ 33^th integral term will be at = 0 + (33 $$-$$ 1)8 = 256

So, there should be at least 256 terms.