Question

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

Answer 1

$(2-x+3x^2{)}^6=[2-x(1-3x){]}^6$
$=2^6-{}^6{C}_1{2}^{5}x(1-3x)+{}^6{C}_2{2}^4{x}^2(1-3x{)}^2-{}^6{C}_3{2}^3{x}^3(1-3x{)}^3+{}^6{C}_4{2}^2{x}^4(1-3x{)}^4-{}^6{C}_{5}2\times x^5(1-3x{)}^5+{}^6{C}_6{x}^6(1-3x{)}^6$

Clearly, $x^4$ occurs in $3^{rd},4^{th}$ and $5^{th}$ terms.

Now, $3^{rd}$ term $=15\times 16x^2(1-6x+9x^2)$
Here, the coefficient of $x^4$ is $15\times 16\times 9=2160.$

$4^{th}$ term $=20\times 8x^3(1-9x+27x^2-27x^3)$
Here, the coefficient of $x^4$ is $20\times 8\times 9=1440$

$5^{th}$ term $=15\times 4x^4[1-4\times 3x+...+(3x{)}^4]$
Here, the coefficient of $x^4$ is $15\times 4=60$

Hence, the required coefficient of $x^4=2160+1440+60=3660$