Question

Find the term of the expansion of ${(\frac{1}{\sqrt[3]{3a^2}}+\sqrt[4]{4a^3})}^{17}$ which does not contain $a$.

• A. $T_3$
• B. $T_5$
• C. $T_7$
• D. $T_9$

Answer 1

The correct option is D $T_9$

Note:- The general term or $(r+1{)}^{th}$ term of $(a+b{)}^n$ is $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

$T_{r+1}{=}^7{C}_r{\left(\frac{1}{\sqrt[3]{3a^2}}\right)}^{17-r}{\left(\sqrt[4]{4a^3}\right)}^r$

$\Rightarrow T_{r+1}{=}^7{C}_r\frac{1}{(a^2{)}^{17-r/3}}{a^3}^{r/4}$

$\Rightarrow T_{r+1}{=}^7{C}_r{a}^{3r/4-2/3(17-r)}$

Now, if power of $x$ must be $0$, then $r$ is:-

$\therefore a^{(3r/4-34/3+2r/3)}=a^0$

$\Rightarrow \frac{3r}{4}-\frac{34}{3}+\frac{2}{3}r=0$

$\Rightarrow \frac{9r+8r}{412}=\frac{34}{3}$

$\Rightarrow 17r=34\times 4$

$\Rightarrow r=8$

Therefore $T_{8+1}=T_9$ is the term which does not contain $a$