The constant term in the expansion of left(fr | vedclass

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Question

(C) Let the expression be $E = \left(\frac{x}{2} + \frac{1}{x} + \sqrt{2}\right)^5$. We can rewrite the expression as: $E = \left(\frac{x^2 + 2 + 2\sq

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(C) Let the expression be $E = \left(\frac{x}{2} + \frac{1}{x} + \sqrt{2}\right)^5$. We can rewrite the expression as: $E = \left(\frac{x^2 + 2 + 2\sqrt{2}x}{2x}\right)^5 = \frac{(x + \sqrt{2})^{10}}{32x^5}$. The constant term in the expansion of $E$ is the coefficient of $x^5$ in the expansion of $(x + \sqrt{2})^{10}$ divided by $32$. Using the Binomial Theorem,the general term in $(x + \sqrt{2})^{10}$ is $T_{r+1} = {}^{10}C_r \cdot x^{10-r} \cdot (\sqrt{2})^r$. For the coefficient of $x^5$,we set $10-r = 5$,which gives $r = 5$. The term is ${}^{10}C_5 \cdot (\sqrt{2})^5 = 252 \cdot 4\sqrt{2} = 1008\sqrt{2}$. Thus,the constant term is $\frac{1008\sqrt{2}}{32} = \frac{63\sqrt{2}}{2}$. Comparing this with $\frac{a\sqrt{2}}{2}$,we get $a = 63$.

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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