If the term independent of x in the expansion of (ax^2+12x^3)^10 is 105 , then a^2 is equa…

Question

If the term independent of $x$ in the expansion of ${(\sqrt{a}{x}^{2}+\frac{1}{2{x}^{3}})}^{10}$ is 105 , then ${a}^{2}$ is equal to :

Accepted Answer

$\begin{gathered} {(\sqrt{a}{x}^{2}+\frac{1}{2{x}^{3}})}^{10} \ {T}_{r+1}={}^{10}{C}_{r}{(\sqrt{a}{x}^{2})}^{10-r}{(\frac{1}{2{x}^{3}})}^{r} \end{gathered}$

Independent of $x\Rightarrow 20-2r-3r=0$

$r=4$

Independent of $x$ is ${}^{10}{C}_{4}(\sqrt{a}{)}^{6}{(\frac{1}{2})}^{4}=105$

$\begin{gathered} \frac{210}{2	imes 8}{a}^{3}=105 \ \Rightarrow \, a=2 \ {a}^{2}=4 \end{gathered}$

Solutions · 1