Question

Find the positive value of $a$ so that the coefficient of $x^5$ is equal to that of $x^{15}$ in the expansion of ${(x^2+\frac{a}{x^3})}^{10}$.

Answer 1

${(x^2+\frac{a}{x^3})}^{10}$

As we know that the general term $\left(T_{r+1}\right)$ of the expression ${(a+b)}^n$ is-

$T_{r+1}={}_n{C}_r{a}^{n-r}{b}^n$

Therefore, in the given expansion ${(x^2+\frac{a}{x^3})}^{10}$,

$a=x^2$

$b=\frac{a}{x^3}$

$n=10$

$\therefore$ general term of the given expression-

$T_{r+1}={}_{10}{C}_r{x}^{20-2r}\frac{a^r}{x^{3r}}$

$\Rightarrow T_{r+1}={}_{10}{C}_r{x}^{20-5r}{a}^r$

For coefficient of $x^5$-

$20-5r=5$

$5r=15\Rightarrow r=3$

$\therefore T_4={}_{10}{C}_3{x}^5{a}^3$

For coefficient of $x^{15}$-

$20-5r=15$

$5r=5\Rightarrow r=1$

$\therefore T_2={}_{10}{C}_1{x}^{15}a$

Given that the coefficient of $x^5$ is equal to the coefficient of $x^{15}$, i.e.,

${}_{10}{C}_3{a}^3={}_{10}{C}_{1}a.....\left(1\right)$

As we know that,

${}_n{C}_r=\frac{n!}{r!(n-r)!}$

$\therefore {}_{10}{C}_3=\frac{10!}{3!(10-3)!}=120$

${}_{10}{C}_1=\frac{10!}{1!(10-1)!}=10$

Now from ${eq}^n\left(1\right)$, we have

$120\times a^3=10a$

$\Rightarrow a^2=\frac{1}{12}$

$\Rightarrow a=\pm \frac{1}{2\sqrt{3}}$

As we have to find positive value of $a$.

$\therefore a=\frac{1}{2\sqrt{3}}$