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Q. The sum of the coefficients of all the even powers of $x$ in the expansion of $\left(2 x^{2}-3 x+1\right)^{11}$ is

Binomial Theorem

Solution:

Let $\left(2 x^{2}-3 x+1\right)^{11}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{22} x^{22}$
to find $a_{0}+a_{2}+a_{4}+\cdots+a_{22}$, we find that
$S_{E}+S_{0}=P(1)=0$ where $P(x)=\left(2 x^{2}-3 x+1\right)^{11}$
$S_{E}-S_{0}=P(-1)=6^{11} \Rightarrow 2 S_{E}=6^{11} \Rightarrow S_{E}=3 \times 6^{10}$