Finding the coefficient of $x^{10}$

Question

I'm doing my maths homework and can't do one of the questions on the binomial expansion. The question is:

a) Find the first four terms in the expansion of $(1-x^2)^8$

b) Hence find the coefficient of $x^{10}$ in the expansion of $(1-2x)(1-x^2)^8$

I can do the first part of the question fine ($1-8x^2+28x^4-56x^6+...$) but am unsure how to do part (b). Can you help me?

Answer 1

I suspect a) actually wants the coefficients of $(x^2)^8 +\ldots + (x^2)^5$. Then b) should be straightforward noticing that all other terms can't contribute to the $x^{10}$.
Name $p(x) = (1-x^2)^8 = a_{16} x^{16} + a_{14} x^{14} + \ldots$ then $$(1-2x)p(x) = p(x) - 2xp(x) = \ldots + a_{10} x^{10} - 2 xa_9 x^9 + \ldots = (a_{10} -2a_9) x^{10} + \text{rest}$$ Now note that $p(x)$ has no $x^9$ term, hence $a_9 = 0$ so b) is no more calculations, just working with a).

Answer 2

The coefficient of $x^{10}$ in the expansion of $(1-2x)(1-x^2)^8$

$1\cdot$ the coefficient of $x^{10}$ in the expansion of $(1-x^2)^8$

$-2\cdot$the coefficient of $x^9$ in the expansion of $(1-x^2)^8$

Clearly, $(1-x^2)^8$ does not have any odd power of $x$ in the binomial expansion

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align} \pars{1 - x^{2}}^{8}&=\sum_{k\ =\ 0}^{8}{8 \choose k}\pars{-1}^{k}x^{2k} \end{align} The first four terms are given by \begin{align} &{8 \choose 0}\pars{-1}^{0}x^{2\times 0} +{8 \choose 1}\pars{-1}^{1}x^{2\times 1}+{8 \choose 2}\pars{-1}^{2}x^{2\times 2} +{8 \choose 3}\pars{-1}^{3}x^{2\times 3} \\[5mm]&=\color{#66f}{\large 1 - 8x^{2} + 28x^{4} - 56x^{6}} \end{align}

\begin{align} \pars{1 - 2x}\pars{1 - x^{2}}^{8}& =\pars{1 - 2x}\sum_{k\ =\ 0}^{8}{8 \choose k}\pars{-1}^{k}x^{2k} \\[5mm]&=\sum_{k\ =\ 0}^{8}{8 \choose k}\pars{-1}^{k}\pars{x^{2k} - 2x^{2k + 1}} \end{align} Clearly, the coefficient of $\ds{x^{10}}$ in the expansion of $\ds{\pars{1 - 2x}\pars{1 - x^{2}}^{8}}$ is given by $$ {8 \choose 5}\pars{-1}^{5}=\color{#66f}{\large -56} $$