Question

The coefficient of $x^{11}$ in the expansion $(1+3x+2x^2{)}^6$ is

• A. $144$
• B. $576$
• C. $288$
• D. $416$

Answer 1

The correct option is B $576$

The general term in the expansion of $(1+3x+2x^2{)}^6$ is
$=\sum \frac{6!}{(r!)(s!)(t!)}\left(1{)}^r\right(3x{)}^s(2x^2{)}^t$
$=\sum \frac{6!}{(r!)(s!)(t!)}\left(1{)}^r\right(3{)}^s\left(2{)}^t\right(x{)}^{s+2t}$

where, $r+s+t=6$
$0\le r,s,t\le 6$

To find the coefficient of $x^{11}$, we need to compare $x^{11}$ with the general term in the expansion of $(1+3x+2x^2{)}^6$ i.e. $x^{11}\to (x{)}^{s+2t}$
$\Rightarrow 11=s+2t$ and $r+s+t=6$
Substitute $s=11-2t$ in $r+s+t=6,$ we get,
$r=t-5$
$\Rightarrow r=0,t=5,s=1$

$\therefore$ Coefficient of $x^{11}$
$=\frac{6!}{(r!)(s!)(t!)}\left(1{)}^r\right(3{)}^s(2{)}^t$
$=\frac{6!}{(0!)(1!)(5!)}\left(1{)}^0\right(3{)}^1(2{)}^5$
$=6\times 3\times 2^5$
$=576$