Question

Find the term independent of x in the expansion of $(x^2+2x^4)(x+\frac{1}{x}{)}^{32}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$(x^2+2x^4)(x+\frac{1}{x}{)}^{32}$ can be written as $\left(x^2\right)(x+\frac{1}{x}{)}^{32}+(2x^4\left)\right(x+\frac{1}{x}{)}^{32}$

To find the term independent of x, we find the terms independent of x in $\left(x^2\right)(x+\frac{1}{x}{)}^{32}$ and $\left(2x^4\right)(x+\frac{1}{x}{)}^{32}$ and add them.

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 ${}^{32}{\text{C}}_{17}.................\left(1\right)$

Similarly for $\left(2x^4\right)(x+\frac{1}{x}{)}^{32}$, we have to find a term which has the power of x equal to -4 in $(x+\frac{1}{x}{)}^{32}$ to get a term independent of x in the expansion $\left(2x^4\right)(x+\frac{1}{x}{)}^{32}$.

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

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

We want the power of x to be -4

$\Rightarrow$ 32-2r = -4

$\Rightarrow$ 2r = 36

$\Rightarrow$ r = 18

$\Rightarrow$ The term independent of x in the expansion of $\left(2x^4\right)(x+\frac{1}{x}{)}^{32}$ is $2^{32}{\text{C}}_{18}...............\left(2\right)$

Combining (1) and (2), we get,

${}^{32}{\text{C}}_{17}+2^{32}{\text{C}}_{18}$