Question

If the number of integral terms in the expansion of ${\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)}^n$ is exactly $33$, then the least value of n is

• A. $128$
• B. $248$
• C. $256$
• D. $264$

Answer 1

The correct option is C

$256$

Explanation for correct option:

In the binomial expansion of ${\left(a+b\right)}^n$ the $r+1$ term is defined as $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$

Therefore, for the expansion of ${\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)}^n$ the $r+1$ is $T_{r+1}={}^{n}C_r{\left(3^{\frac{1}{2}}\right)}^{n-r}{\left(5^{\frac{1}{8}}\right)}^r$

$\therefore 3^{\frac{n-r}{2}}$ will be integer for $n-r=0,2,4,.....$ and $5^{\frac{r}{8}}$ will be integer for $r=0,8,16,24$

So $3^{\frac{n-r}{2}}$and $5^{\frac{r}{8}}$ will have integer $0,8,16,24,......$

Therefore the common difference is $8$

Let $a_0=0$

There are $33$ term with an integer term,

So $n=33$

The $n^{th}$ term is given by

$$\begin{gathered} a_n=a_0+d\left(n-1\right) \\ =0+8\left(33-1\right) \\ =256 \end{gathered}$$

Hence, option (C) $256$ is the correct answer.