First slide

Binomial theorem for positive integral Index

Question

Coefficient of the term independent of x in the expansion ${(x + \frac{1}{x})}^{4} {(x - \frac{1}{x})}^{12}$ is

Moderate

A

192
B

194
C

196
D

198

Solution

$\begin{array}{l} {(x + \frac{1}{x})}^{4} {(x - \frac{1}{x})}^{12} = {(x + \frac{1}{x})}^{4} {(x - \frac{1}{x})}^{4} {(x - \frac{1}{x})}^{8} \\ = {(x^{2} - \frac{1}{x^{2}})}^{4} {(x - \frac{1}{x})}^{8} \\ = (^{4} C_{0} x^{8} -^{4} C_{1} x^{4} +^{4} C_{2} -^{4} C_{3} \frac{1}{x^{4}} +^{4} C_{4} \frac{1}{x^{8}}) \times (^{8} C_{0} x^{8} -^{8} C_{1} x^{6} +^{8} C_{2} x^{4} - \dots +^{8} C_{8} \frac{1}{x^{8}}) \end{array}$
$∴$ Coefficient of the term independent of x
$\begin{aligned} = (^{4} C_{0}) (^{8} C_{8}) - (^{4} C_{1}) (^{8} C_{6}) + (^{4} C_{2}) (^{8} C_{4}) - (^{4} C_{3}) (^{8} C_{2}) + (^{4} C_{4}) (^{8} C_{0}) = 198 \end{aligned}$

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