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Q.
If $\left(1+x+x^2+x^3\right)^5=a_0+a_1 x+a_2 x^2+\ldots \ldots+a_{15} x^{15}$, then $a_{10}$ is equal to
Binomial Theorem
Solution:
Coefficient of $x ^{10} \text { in }\left(1+ x + x ^2+ x ^3\right)^5$
$a _{10}=\left(1+ x + x ^2(1+ x )\right)^5=(1+ x )^5\left(1+ x ^2\right)^5$
$=\left({ }^5 C _0+{ }^5 C _1 x +{ }^5 C _2 x ^2+{ }^5 C _3 x ^3+{ }^5 C _4 x ^4+{ }^5 C _5 x ^5\right)\left({ }^5 C _0+{ }^5 C _1 x ^2+{ }^5 C _2 x ^4+{ }^5 C _3 x ^6+{ }^5 C _4 x ^8+{ }^5 C _5 x ^{10}\right) $
$={ }^5 C _0{ }^5 C _5+{ }^5 C _2{ }^5 C _4+{ }^5 C _4{ }^5 C _3=1+10 \cdot 5+5 \cdot 10 $
$ a _{10}=101 $