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. 
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\pars{x^{2} + 2x + 2}^{10} =
\sum_{a,b,c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose a,b,c}
\bracks{a + b + c = 10}x^{2a}\,\pars{2x}^{b}\,2^{c}
\\[5mm] = &\
\sum_{a,b,c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose a,b,c}2^{b + c}
\bracks{a + b + c = 10}\sum_{n = 0}^{\infty}\bracks{n = 2a + b}x^{n}
\end{align}

$$
\bracks{x^{n}}\pars{x^{2} + 2x + 2}^{10} =
\sum_{a,b,c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose a,b,c}2^{b + c}
\bracks{a + b + c = 10}\bracks{n = 2a + b}
$$
The sum restrictions let to write $\ds{a}$ and $\ds{b}$ in terms of $\ds{c}$ and $\ds{n}$. Namely,
$\ds{a = n + c - 10}$ and $\ds{b = 20 - 2c - n}$:
$$
\bracks{x^{n}}\pars{x^{2} + 2x + 2}^{10} =
\sum_{c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose n + c - 10,20 - 2c - n,c}
2^{20 - c - n}
$$
$\ds{c}$ bounds are determined by:
$$
a = n + c - 10 \geq 0\,,\quad b = 20 - 2c - n \geq 0
\implies 10 - n \leq c \leq 10 - n/2
$$

\begin{align}
&\bracks{x^{n}}\pars{x^{2} + 2x + 2}^{10} =
\sum_{c = m}^{M}{10! \over \pars{n + c - 10}!\pars{20 - c - n}!c!}\,
2^{20 - c - n}
\\[5mm] = &\,\,\,
\bbox[#ffe,20px,border:1px dotted navy]{\ds{\left. 2^{20 - n}\sum_{c = m}^{\left\lfloor 10 - n/2\right\rfloor}
{10 \choose c}{10 - c \choose n + c - 10}2^{-c}\right\vert_{\ n\ \leq\ 20}}}
\qquad\mbox{where}\qquad m = \max\braces{0,10 - n}
\end{align}
Note that the above expression vanishes out when
$\ds{\quad n <0\quad \mbox{o}\quad n > 20}$.


For instance: When $\ds{n = 3}$,

\begin{align}
&\bracks{x^{3}}\pars{x^{2} + 2x + 2}^{10} =
2^{17}\sum_{c = 7}^{8}
{10 \choose c}{10 - c \choose c - 7}2^{-c} =
2^{17}\bracks{{10 \choose 7}2^{-7} + 2{10 \choose 8}2^{-8}}
\\[5mm] & =
120 \times 2^{10} + 45 \times 2^{10} = \bbx{168960}
\end{align}
