Question

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

Answer 1

$(2-x+3x^2{)}^6=\sum _{r=0}^6{}^6{C}_r(2-x{)}^{6-r}(3x^2{)}^r=\sum _{r=0}^6{}^6{C}_r(2-x{)}^{6-r}{3}^r{x}^{2r}$
Clearly, $x^4$ occurs for $r=0,1,2$

For $r=0$
${}^6{C}_0(2-x{)}^6$
Then, coefficient of $x^4$ in this
$={}^6{C}_0\cdot {}^6{C}_4\cdot (2{)}^2=60$
For $r=1$
${}^6{C}_1(2-x{)}^5\cdot 3\cdot x^2$
Then, coefficient of $x^4$ in this
$={}^6{C}_1\cdot {}^5{C}_2\cdot (2{)}^3\cdot 3=1440$
For $r=2$
${}^6{C}_2(2-x{)}^4\cdot 3^2\cdot x^4$
Then, coefficient of $x^4$ in this
$={}^6{C}_2\cdot {}^4{C}_0\cdot (2{)}^4\cdot 3^2=2160$
Therefore, the required coefficient of $x^4$
$=60+1440+2160=3660$


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

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 $x^4=2160+1440+60=3660$