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Q.
Find the sum of the coefficients of even powers of $x$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{5}$.
Binomial Theorem
Solution:
$\left(1+x+x^{2}+x^{3}\right)^{5}=\left[(1+x)\left(1+x^{2}\right)\right]^{5}$
$=(1+x)^{5}\left(1+x^{2}\right)^{5}$
$=\left({ }^{5} C _{0}+{ }^{5} C _{1} x+{ }^{5} C _{2} x^{2}+{ }^{5} C _{3} x^{3}+{ }^{5} C _{4} x^{4}+{ }^{5} C _{5} x^{5}\right)$
$ \times\left({ }^{5} C _{0}+{ }^{5} C _{1} x^{2}+{ }^{5} C _{2}\left(x^{2}\right)^{2}+{ }^{5} C _{3}\left(x^{2}\right)^{3}\right.\left.+{ }^{5} C _{4}\left(x^{2}\right)^{4}+{ }^{5} C _{5}\left(x^{2}\right)^{5}\right)$
$=\left(1+5 x+10 x^{2}+10 x^{3}+5 x^{4}+x^{5}\right)$
$\times\left(1+5 x^{2}+10 x^{4}+10 x^{6}+5 x^{8}+x^{10}\right)$
Sum of coefficients $=2^{5}(1+5+10)$
$=32(16)$
$=512$