I understand that the coefficient of, say $x^8$, in the expansion of $(1+x)^{10}$, would be ${10 \choose 8}$, but what about an expression like $(1+x^2)^{10}$? Would I have to square root the ${10 \choose 8}$? And if that is true, would I triple root it for expressions such as $(1+x^2)^{10}$ (if it is even possible)?
1 Answer
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Let $x^2=y$. Then, by the Binomial theorem
$$(1+x^2)^{10}=(1+y)^{10}=\sum_{i=1}^{10}{10\choose i}y^i=\sum_{i=1}^{10}{10\choose i}(x^2)^i=\sum_{i=1}^{10}{10\choose i}x^{2i}$$ So, the coefficient of $x^8$ is ${10\choose 4}$
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$\begingroup$ @StevenGregory Yes, I just fixed that. :) $\endgroup$ Jun 5, 2016 at 1:38