Question

The ratio of the co-efficient of $x^{12}$ to the term independent of x in the expansion of ${(x^2+\frac{2}{x})}^{15}$ is

• A. 7 : 64
• B. 64 : 7
• C. 5 : 48
• D. 48 : 5

Answer 1

The correct option is C 5 : 48

Given ${(x^2+\frac{2}{x})}^{15}$
$T_{r+1}{=}^{15}{C}_r(x^2{)}^{15-r}{\left(\frac{2}{x}\right)}^r$
If $T_{r+1}$ contains $x^{12}$, then $r=\frac{30-12}{3}=\frac{18}{3}=6$
$\therefore$ Co =efficient of $x^{12}{=}^{15}{C}_6{2}^6$
If $T_{r+1}$ contains $x^{\circ }$, (term independent of x), then $r=\frac{30}{3}=10$
$\therefore$ Independent of x term ${=}^{15}{C}_{10}{2}^{10}$
$\therefore$ ratio =$\frac{{}^{15}{C}_6{2}^6}{{}^{15}{C}_{10}{2}^{10}}=\frac{{}^{15}{C}_6}{{}^{15}{C}_5}.\frac{1}{2^4}=\frac{5}{3}.\frac{1}{2^4}=\frac{5}{48}$