Question

In the binomial expansion of ${(\sqrt{y}+\frac{1}{2\sqrt[4]{4y}})}^n,$ the first three coefficients form an arithmetic progression. Then

• A. the value of $n$ is $7$
• B. the value of $n$ is $8$
• C. number of terms in the expansion where the power of $y$ is natural number, is $2$
• D. number of terms in the expansion where the power of $y$ is natural number, is $3$

Answer 1

Answer: (B), (C)

The correct options are
B the value of $n$ is $8$
C number of terms in the expansion where the power of $y$ is natural number, is $2$

${(\sqrt{y}+\frac{1}{2\sqrt[4]{4y}})}^n$
First three coefficients ${}^n{C}_0,\frac{{}^n{C}_1}{2},\frac{{}^n{C}_2}{2^2}$ are in A.P.
$\Rightarrow 2\cdot \frac{{}^n{C}_1}{2}={}^n{C}_0+\frac{{}^n{C}_2}{2^2}$
$\Rightarrow n=1+\frac{1}{4}\frac{n!}{2!(n-2)!}$
$\Rightarrow n=1+\frac{1}{4}\frac{n(n-1)}{2}$
$\Rightarrow 8n=8+n^2-n$
$\Rightarrow n^2-9n+8=0$
$\Rightarrow n=8$ or $n=1$ $(n=1$ is rejected)
$\therefore n=8$

The expansion is ${(y^{\frac{1}{2}}+\frac{1}{2}{y}^{\frac{-1}{4}})}^8,$
where $T_{r+1}=\frac{{}^8{C}_r}{2^r}\cdot y^{\frac{8-r}{2}}\cdot y^{\frac{-r}{4}}$
$=\frac{{}^8{C}_r}{2^r}\cdot y^{\frac{16-3r}{4}}$
The terms where power of $y$ is natural are
${}^8{C}_0\cdot y^4\to$ First term where $r=0$
$\frac{{}^8{C}_4\cdot y^1}{2^4}\to$ Fifth term where $r=4$