Question

In the binomial expansion of $\left(\sqrt{ x }+\frac{1}{2 \cdot \sqrt[4]{ x }}\right)^{ n }$, the first three coefficients form an arithmetic progression, then sum of coefficients of all the terms is

• A. $\left(\frac{3}{2}\right)^5$
• B. $\left(\frac{3}{2}\right)^6$
• C. $\left(\frac{3}{2}\right)^7$
• D. $\left(\frac{3}{2}\right)^8$

Answer 1

Answer: (D)

${ }^{ n } C _0+{ }^{ n } C _2 \frac{1}{2^2}=2{ }^{ n } C _1 \frac{1}{2}$
$\Rightarrow 1+\frac{ n ( n -1)}{8}= n =\frac{ n ^2- n +8}{8}$
$\Rightarrow n ^2-9 n +8=0 \Rightarrow n =1,8 n =1 \text { (rejected) }$
$\therefore \text { sum of coefficient }=\left(1+\frac{1}{2}\right)^8=\left(\frac{3}{2}\right)^8 $