Question

Find the coefficient of $a^3$ in the expansion of ${(\sqrt{a}+\frac{1}{a\sqrt{a}})}^{22}$

• A. ${}^{22}{C}_4$
• B. ${}^{22}{C}_6$
• C. ${}^{22}{C}_8$
• D. ${}^{22}{C}_{10}$

Answer 1

The correct option is A ${}^{22}{C}_4$

Given, ${(\sqrt{a}+\frac{1}{a\sqrt{a}})}^{22}$

General term

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^{22}{C}_r(\sqrt{a}{)}^{22-r}{\left(\frac{1}{a\sqrt{a}}\right)}^r$

$T_{r+1}={}^{22}{C}_r(a{)}^{n-\frac{r}{2}}\frac{(1{)}^r}{a^{\frac{3}{2}r}}$

$T_{r+1}={}^{22}{C}_r\left(1{)}^r\right(a{)}^{11-2r}$

Required, coefficient of $a^3$

$11-2r=3$

$11-3=2r$

$2r=8$

$r=4$

So coefficient is ${}^{22}{C}_4$

Hence correct option is A.