Question

The coefficient of $x^{160}$ in the expansion of $(x^8+1{)}^{60}{(x^{12}+3x^4+\frac{3}{x^4}+\frac{1}{x^{12}})}^{-10}$ is

• A. ${}^{30}{C}_6$
• B. ${}^{30}{C}_5$
• C. divisible by 189
• D. divisible by 203

Answer 1

The correct option is B ${}^{30}{C}_5$

$S={(1+x^8)}^{60}.{(x^{12}+{3x}^4+\frac{3}{x^4}+\frac{1}{x^{12}})}^{-10}$

$=x^{12}+{3x}^4+\frac{3}{x^4}+\frac{1}{x^{12}}={(x^4+\frac{1}{x^4})}^3$

$\Rightarrow 3={(1+x^8)}^{60}.{\left\{{\left(\frac{x^8+1}{x^4}\right)}^3\right\}}^{-10}$

$=\frac{{(1+x^8)}^{60}.{(1+x^8)}^{-30}}{x^{-120}}$

$={(1+x^8)}^{30}.x^{120}$

$=[1+{{}^{30}C}_1.x^8+.....{{}^{30}C}_5.{(x}^8{)}^5].x^{120}$

$\therefore$ Coefficient of $x^{160}={{}^{30}C}_5.$

Hence the answer is ${{}^{30}C}_5.$