Find the series expansion of 2 multiplied functions 
The first three terms in the series expansion of $(1+x)^m$ are $1 + mx + \dfrac{m(m-1)x^2}{2}$.
Find the first 3 terms in the series expansion of $(1+x)^{m+1}(1-2x)^m$.

I don't really know how to do this, because I have always done these binomial expansion using binomial coefficients and Pascal's triangle.
 A: The best way is to change your variables around so you don't get confused between $x$'s, $m$'s, and other variables. I recommend rewriting your original expansion as follows:
$${(1+x)}^m \implies {(1+u)}^t = 1 + tu + \frac{t(t-1)u^2}{2}$$
Then think of the second expression as using the arguments with particular values of $t$ and $u$:
$${(1+x)}^{m+1}{(1-2x)}^m \implies {(1+(x))}^{m+1}{(1+(-2x))}^{m} \\ = {(1+u_1)}^{t_1}{(1+u_2)}^{t_2}$$ where $u_1 = x; \quad t_1 = m+1; \quad u_2 = -2x; \quad t_2 = m$.
Then you can expand the expression above, using the first expansion, for the subscripted $u$'s and $t$'s to get:
$${(1+u_1)}^{t_1} = 1 + t_1u_1 + \frac{t_1[t_1-1]{u_1}^2}{2} \\ = 1 + (m+1)(x) + \frac{(m+1)[(m+1)-1](x)^2}{2} \\ = 1 + mx + x + \frac{(m+1)(m)x^2}{2}$$
And similarly for the second expression:
$${(1+u_2)}^{t_2} = 1 + t_2u_2 + \frac{t_2[t_2-1]{u_2}^2}{2} \\ = 1+ (m)(-2x) + \frac{(m)[(m)-1]{(-2x)}^2}{2} \\ = 1-2mx+\frac{m(m-1)4x^2}{2} \\ = 1-2mx+2m(m-1)x^2$$
Now all that needs to be done is multiply the two halves of the original product ${(1+u_1)}^{t_1}{(1+u_2)}^{t_2}$: 
$${(1+u_1)}^{t_1}{(1+u_2)}^{t_2} = \left( 1 + mx + x + \frac{(m+1)(m)x^2}{2} \right) \left( 1-2mx+2m(m-1)x^2 \right)$$
At this point, the product is just tedium to find the first 3 terms of the expansion you're looking for.
A: You can use the Taylor series technique if you are familiar with it. Here are the three desired terms
$$ 1+ \left( 1-m \right) x + \frac{m}{2}( m-7) \,{x}^{2}.$$
A: $\newcommand{\+}{^{\dagger}}%
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$\large\tt Hint:$
\begin{align}
&\color{#0000ff}{\large\pars{1 + x}^{m + 1}\pars{1 - 2x}^{m}}
=
\sum_{\ell = 0}^{m + 1}{m + 1 \choose \ell}x^{\ell}
\sum_{\ell' = 0}^{m}{m \choose \ell'}\pars{-1}^{\ell'}2^{\ell'}x^{\ell'}
\\[3mm]&=
\sum_{\ell = 0}^{m + 1}{m + 1 \choose \ell}x^{\ell}
\sum_{\ell' = 0}^{m}{m \choose \ell'}\pars{-1}^{\ell'}2^{\ell'}
\sum_{n = 0}^{2m + 1}x^{n - \ell}\delta_{n - \ell,\ell'}
\\[3mm]&=
\sum_{n = 0}^{2m + 1}x^{n}
\sum_{\ell = 0}^{m + 1}{m + 1 \choose \ell}\pars{-1}^{n - \ell}2^{n - \ell}
{m \choose n - \ell}\sum_{\ell' = 0}^{m}
\delta_{\ell',n - \ell}
\\[3mm]&=
\sum_{n = 0}^{2m + 1}x^{n}\bracks{%
\sum_{{\vphantom{\Large A}\ell\ =\ 0} \atop {\vphantom{\LARGE A}0\ \leq\ n - \ell\ \leq\ m}}^{m + 1}{m + 1 \choose \ell}{m \choose n - \ell}\pars{-1}^{n - \ell}
2^{n - \ell}}
=
\color{#0000ff}{\large\sum_{n = 0}^{2m + 1}a_{n}x^{n}}
\end{align}

$$
a_{n}
\equiv
\sum_{{\vphantom{\Large A}\ell\ =\ 0}
      \atop
      {\vphantom{\LARGE A}n - m\ \leq\ \ell\ \leq\ n}}^{m + 1}
{m + 1 \choose \ell}{m \choose n - \ell}\pars{-2}^{n - \ell}
$$

\begin{align}
a_{0}
&=
\sum_{{\vphantom{\Large A}\ell\ =\ 0}
      \atop
      {\vphantom{\LARGE A}-m\ \leq\ \ell\ \leq\ 0}}^{m + 1}
{m + 1 \choose \ell}{m \choose 0 -\ell}\pars{-2}^{-\ell} = 1
\\[3mm]
a_{1}
&=
\sum_{{\vphantom{\Large A}\ell\ =\ 0}
      \atop
      {\vphantom{\LARGE A}1 - m\ \leq\ \ell\ \leq\ 1}}^{m + 1}
{m + 1 \choose \ell}{m \choose 1 - \ell}\pars{-2}^{1 - \ell}
\\[3mm]
a_{2}
&=
\sum_{{\vphantom{\Large A}\ell\ =\ 0}
      \atop
      {\vphantom{\LARGE A}2 - m\ \leq\ \ell\ \leq\ 2}}^{m + 1}
{m + 1 \choose \ell}{m \choose 2 - \ell}\pars{-2}^{2 - \ell}
\end{align}
