Question

The coefficient of $x^5$ in the expansion of $(2-x+3x^2{)}^6$ is

• A. - 4692
• B. 4692
• C. 2346
• D. - 5052

Answer 1

Answer: (D)

The general term in the expansion of
${\left(2-x+3x^2\right)}^6$ is
$\frac{6!}{r!s!t!}{2}^r(-x{)}^s{\left(3x^2\right)}^t$, where $r+s+t=6$
$=\frac{6!}{r!s!t!}{2}^r\times (-1{)}^s\times 3^t\times x^{s+2t}$
where $r+s+t=6$
For the coefficient of $x^5$, we must have
$s+2t=5$
But $r+s+t=6$
$\therefore s=5-2t$ and $r=1+t$
where, $0\le r,s,t\le 6$
Now, $t=0\Rightarrow r=1,s=5$
$t=1\Rightarrow r=2,s=3$
$t=2\Rightarrow r=3,s=1$
Thus, there are three terms containing $x^5$ and coefficient of $x^5$
$=\frac{6!}{1!5!0!}\times 2^1\times (-1{)}^5\times 3+\frac{6!}{2!3!1!}\times 2^2$
$\times (-1{)}^3\times 3^1+\frac{6!}{3!1!2!}\times 2^3\times (-1{)}^1\times 3^2$
$=-12-720-4320$
$=-5052$