Question

Consider the binomial expansion of ${(\sqrt{x}+\frac{1}{2\cdot \sqrt[4]{4x}})}^n,n\in N$ where the terms of the expansion are written in decreasing powers of $x.$ If the coefficients of the first three terms form an arithmetic progression, then which of the following statements hold(s) good?

• A. Total number of terms in the binomial expansion is $8$
• B. Number of terms in the binomial expansion with integral power of $x$ is $3$
• C. There is no term in the binomial expansion which is independent of $x$
• D. Fourth and fifth terms are the middle terms of the expansion

Answer 1

Answer: (B), (C)

There is no term in the binomial expansion which is independent of $x$

${(x^{1/2}+\frac{1}{2}{x}^{-1/4})}^n$
$T_{r+1}={}^n{C}_r{x}^{(n-r)/2}\cdot \frac{1}{2^r}\cdot x^{-r/4}$
Coefficient of the first three terms are
${}^n{C}_0,{}^n{C}_1\cdot \frac{1}{2},{}^n{C}_2\cdot \frac{1}{2^2}\therefore {}^n{C}_0+{}^n{C}_2\cdot \frac{1}{4}=2\cdot {}^n{C}_1\cdot \frac{1}{2}\Rightarrow 1+\frac{n(n-1)}{8}=n\Rightarrow \frac{n(n-1)}{8}=n-1\Rightarrow n=8(\because n\ne 1)$

$\therefore T_{r+1}={}^n{C}_r{x}^{(n-r)/2}\cdot \frac{1}{2^r}\cdot x^{-r/4}={}^8{C}_r\cdot \frac{1}{2^r}\cdot x^{(4-\frac{3r}{4})}$

Hence, there are total $9$ terms in the binomial expansion.
And $5$th term is the middle term.

Terms with integer power occur when $r=0,4,8$
Hence, $3$ such terms are there.

For being independent of $x:$
$4-\frac{3r}{4}=0$
$\Rightarrow r=\frac{16}{3}$ (not possible)
Hence, no such term is there.