Question

The coefficient of $x^8$ in the expansion of ${\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}\right)}^2$ is

• A. $\frac{1}{315}$
• B. $\frac{2}{315}$
• C. $\frac{1}{105}$
• D. $\frac{1}{210}$

Answer 1

Answer: (A)

Coefficient of $x^8$ in
$\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}\right)\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}\right)$
$=1\times \frac{1}{8!}+\frac{1}{2!}\times \frac{1}{6!}+\frac{1}{4!}\times \frac{1}{4!}+\frac{1}{6!}\times \frac{1}{2!}+\frac{1}{8!}\times 1$
$=\frac{1}{0!8!}+\frac{1}{2!6!}+\frac{1}{4!4!}+\frac{1}{6!2!}+\frac{1}{8!0!}$
$=\frac{1}{8!}\left(\frac{8!}{0!8!}+\frac{8!}{2!6!}+\frac{8!}{4!4!}+\frac{8!}{6!2!}+\frac{8!}{8!0!}\right)$
$=\frac{1}{8!}\left({}^8{C}_0{+}^8{C}_2{+}^8{C}_4{+}^8{C}_6{+}^8{C}_8\right)$
$=\frac{2^{8-1}}{8!}=\frac{2^7}{8!}=\frac{1}{315}$