Question

If $n$ is the number of irrational terms in the expansion of ${[3^{\frac{1}{4}}+5^{\frac{1}{8}}]}^{60},$ then $(n-1)$ is divisible by

• A. $8$
• B. $26$
• C. $7$
• D. $30$

Answer 1

The correct option is B

$26$

Explanation for correct answer:

Binomial exapansion:

We know that for the binomial expansion of ${\left(a+b\right)}^n$ the ${\left(r+1\right)}^{th}$ term is $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Given expression is ${[3^{\frac{1}{4}}+5^{\frac{1}{8}}]}^{60},$

Therefore $T_{r+1}={}^{60}{C}_r{\left(3^{\frac{1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{8}}\right)}^r$

$T_{r+1}={}^{60}{C}_r{\left(3\right)}^{\frac{(60-r)}{4}}{\left(5\right)}^{\frac{r}{8}}$

It is rational if $\frac{[60–r]}{4},\left(\frac{r}{8}\right)$, both are whole numbers, $r\in \{0,1,2,\dots .60\}$

$\Rightarrow \frac{[60–r]}{4}\in W\Rightarrow r\in \{0,4,8,\dots \dots ..60\}$ and ,

$\left(\frac{r}{8}\right)\in W\Rightarrow r\in \{0,8,16,\dots 56\}$

Common terms $r\in \{0,8,16,\dots .56\}$

So $8$ terms are rational

Total terms = $61$

Then irrational terms$=61–8=53=n$

$\therefore n–1=52$

Factors $52$ are $1,2,4,13,26,52.$

Here, $n–1=52$is divisible by $26$

Hence, option B is correct