The coefficient of x^{4} in the expansion of | vedclass

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Question

$\left(1+x+x^{2}+x^{3}\right)^{6}=\left((1+x)\left(1+x^{2}\right)\right)^{6}$

$=(1+x)^{6}\left(1+x^{2}\right)^{6}$

$=\sum_{ r =0}^{6}{ }^{6} C _

Vedclass · Accepted Answer

$120$

Vedclass · Answer

$\left(1+x+x^{2}+x^{3}ight)^{6}=\left((1+x)\left(1+x^{2}ight)ight)^{6}$

$=(1+x)^{6}\left(1+x^{2}ight)^{6}$

$=\sum_{ r =0}^{6}{ }^{6} C _{ r } x ^{ r } \sum_{ r =0}^{6}{ }^{6} C _{ t } x ^{2 t }$

$=\sum_{ r =0}^{6} \sum_{ t =0}^{6}{ }^{6} C _{ r }^{6} C _{ t } x ^{ r +2 t }$

For coefficient of $x^{4} \Rightarrow r+2 t=4$

$\begin{array}{|c|c|} \hline r & t \ \hline 0 & 2 \ \hline 2 & 1 \ \hline 4 & 0 \ \hline \end{array}$

Coefficient of $x^{4}$ $={ }^{6} C _{0}{ }^{6} C _{2}+{ }^{6} C _{2}{ }^{6} C _{1}+{ }^{6} C _{4}{ }^{6} C _{0}$

$=120$

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

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