<s>[INST] Coefficient of a term using binomial theorem I was just wondering how would I find the coefficient of any term let's say $x^3$ in the expansion of $(x^2+2x+2)^{10}$ using binomial expansion or any other technique. Please let me know if this can be found directly using a shortcut if any. 
 [/INST] We can also apply the binomial theorem twice in order to determine the coefficient of $x^3$ in $(x^2+2x+10)^{10}$.
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
[x^3](x^2+2x+2)^{10}&=[x^3]\sum_{k=0}^{10}\binom{10}{k}x^{2k}(2x+2)^{10-k}\tag{1}\\
&=\sum_{k=0}^{1}\binom{10}{k}[x^{3-2k}]\sum_{j=0}^{10-k}\binom{10-k}{j}(2x)^j2^{10-k-j}\tag{2}\\
&=\sum_{k=0}^1\binom{10}{k}\binom{10-k}{3-2k}2^{10-k}\tag{3}\\
&\binom{10}{0}\binom{10}{3}2^{10}+\binom{10}{1}\binom{9}{1}2^9\tag{4}\\
&=1\cdot120\cdot1024+10\cdot9\cdot 512\\
&=168960
\end{align*}

Comment:


*

*In (1) we apply the binomial formula

*In (2) we observe that only $k=0$ and $k=1$ may contribute via $x^{2k}$ something for the coefficient of $x^3$. So, we change the upper limit of the sum to $k=1$. We apply the coefficient of operator rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
and we also expand the inner binomial according to the binomial formula.

*In (3) we select the coefficient of $x^{3-2k}$ by selecting the summand with $j=3-2k$.

*In (4) we expand the sum.
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