Question

Find the coefficient of $x^4$ in the expansion of ${(2-x+{3x}^2)}^6$

• A. $3661$
• B. $1830$
• C. $3300$
• D. none of the above

Answer 1

The correct option is D none of the above

We want the coefficient of $x^4$ in $(2x^0-1x+3x^2{)}^6$ will be in the case $\left(x^0{)}^3\right(x^1{)}^1(x^2{)}^2$ , $\left(x^0{)}^2\right(x^1{)}^4(x^2{)}^0$ and $\left(x^0{)}^4\right(x^1{)}^0(x^2{)}^2$
Coefficient of $x^4$ in:
$\left(x^0{)}^3\right(x^1{)}^1(x^2{)}^2=\frac{6!\left(2{)}^3\right(-1{)}^1(3{)}^2}{3!1!2!}=-4320$
$\left(x^0{)}^2\right(x^1{)}^4(x^2{)}^0=\frac{6!\left(2{)}^2\right(-1{)}^4(3{)}^0}{2!4!0!}=60$
$\left(x^0{)}^4\right(x^1{)}^0(x^2{)}^2=\frac{6!\left(2{)}^4\right(-1{)}^0(3{)}^2}{4!0!2!}=2160$
So, total will be $2160+60-4320=-2100$
i.e., none of the above.