# 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.

– user371838
Nov 4 '16 at 13:45
• Alright, I'll just check it now. Nov 4 '16 at 13:46
• Vieta's formula probably a good idea. Nov 4 '16 at 13:58

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.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}