Question

Find the term independent of x in the expansion of $x^2{(x+\frac{1}{x})}^{32}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Consider $\left(x^2\right){(x+\frac{1}{x})}^{32}$, the term independent of x in this expansion will have a power of x equal to -2 in the expansion of ${(x+\frac{1}{x})}^{32}$ (Then only the power of x in $\left(x^2\right){(x+\frac{1}{x})}^{32}$ will become zero, because there is an $x^2$ term outside)

$T_{r+1}$ = ${}^{32}{\text{C}}_r{x}^{32-r}{\left(\frac{1}{x}\right)}^r$

= ${}^{32}{\text{C}}_r{x}^{32-2r}$

We want the power of x to be -2

$\Rightarrow$ 32 - 2r = -2

$\Rightarrow$ 34 = 2r

$\Rightarrow$ r = 17

$\Rightarrow$ The term independent of x in the expansion of $\left(x^2\right){(x+\frac{1}{x})}^{32}$ is ${}^{32}{\text{C}}_{17}$