Question

Find the coefficient of $x^5$ in ${\left(1+x+x^2\right)}^8$.

• A. 405
• B. 508
• C. 404
• D. 504

Answer 1

Answer: (D)

Given, Expansion is ${\left(1+x+x^2\right)}^8$
Coefficient of $x^5$ in above Expansion is
$\sum \frac{8!}{n_1!n_2!n_3!}(1{)}^{n_1}\cdot (x{)}^{n_2}{\left(x^2\right)}^{n_3}$
$\therefore \left(n_1+n_2+n_3\right)=8$
and $n_2+2n_3=5$

$n_1$	$n_2$	$n_3$
$5$	$3$	$0$
$5$	$1$	$2$
$4$	$3$	$1$

Coefficient of $x^5=\frac{8!}{5!3!0!}+\frac{8!}{5!1!2!}+\frac{8!}{4!3!1!}$
$=56+168+280=504$