The constant term in the expansion of $\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)^5$ is $\frac{a \sqrt{2}}{2}$, then $a=$

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Aicharya · Accepted Answer

$\begin{aligned} & \left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)^5 \ & \Rightarrow \quad\left(\sqrt{\frac{x}{2}}+\frac{1}{\sqrt{x}}\right)^{10}=\left(\frac{x+\sqrt{2}}{\sqrt{2} x}\right)^{10} \Rightarrow \frac{(x+\sqrt{2})^{10}}{32 x^5} \end{aligned}$ Forms constant term in the expansion $\begin{array}{r} =\frac{\text { Coefficient of } x^5 \text { in }(x+\sqrt{2})^{10}}{32} \ \Rightarrow \quad \frac{{ }^{10} C_5 \times 4 \sqrt{2}}{32}=\frac{a \sqrt{2}}{2} \Rightarrow(a=63) \end{array}$

The constant term in the expansion of \(\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)^5\) is \(\frac{a…

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