Question

The coefficient of $x^4$ in the expansion of ${\left(1+x+x^2+x^3\right)}^6$ in powers of $x,$ is_______

Answer 1

Answer: 120.00

${\left(1+x+x^2+x^3\right)}^6={\left((1+x)\left(1+x^2\right)\right)}^6$
$=(1+x{)}^6{\left(1+x^2\right)}^6$
$=\sum _{r=0}^6{}^6{C}_r{x}^r\sum _{r=0}^6{}^6{C}_t{x}^{2t}$
$=\sum _{r=0}^6\sum _{t=0}^6{}^6{C}_r{}^6{C}_t{x}^{r+2t}$
For coefficient of $x^4$
$\Rightarrow r+2t=4$

r	t
0	2
2	1
4	0

Coefficient of $x^4$
$={}^6{C}_0{}^6{C}_2+{}^6{C}_2{}^6{C}_1+{}^6{C}_4{}^6{C}_0$
$=120$