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7.Binomial Theorem
medium
$x$ की घातों में $\left(1+x+x^{2}+x^{3}\right)^{6}$ के प्रसार में $x^{4}$ का गुणांक है .............
A
$116$
B
$118$
C
$120$
D
$124$
(JEE MAIN-2020)
Solution
$\left(1+x+x^{2}+x^{3}\right)^{6}=\left((1+x)\left(1+x^{2}\right)\right)^{6}$
$=(1+x)^{6}\left(1+x^{2}\right)^{6}$
$=\sum_{ r =0}^{6}{ }^{6} C _{ r } x ^{ r } \sum_{ r =0}^{6}{ }^{6} C _{ t } x ^{2 t }$
$=\sum_{ r =0}^{6} \sum_{ t =0}^{6}{ }^{6} C _{ r }^{6} C _{ t } x ^{ r +2 t }$
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 $x^{4}$ $={ }^{6} C _{0}{ }^{6} C _{2}+{ }^{6} C _{2}{ }^{6} C _{1}+{ }^{6} C _{4}{ }^{6} C _{0}$
$=120$
Standard 11
Mathematics
