If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$ is 105 , then $\mathrm{a}^2$ is equal to :
Solution
<p>$$\begin{aligned}
& \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\
& T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r
\end{aligned}$$</p>
<p>Independent of $x \Rightarrow 20-2 r-3 r=0$</p>
<p>$r=4$</p>
<p>Independent of $x$ is ${ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105$</p>
<p>$$\begin{gathered}
\frac{210}{2 \times 8} a^3=105 \\
\Rightarrow \quad a=2 \\
a^2=4
\end{gathered}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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