I'm attempting this question on binomial distributions:

$\displaystyle 1 + mx + \frac{m(m-1)x^2}{2} $ are the first 3 terms of the series expansion of $\displaystyle (1+x)^m $

Use this to find the 1st 3 terms of the series expansion of $\displaystyle (1+x)^{m+1} (1-2x)^m $

This is what I've done:

1) Substituted (m+1) into the initial expansion to give the expansion for $\displaystyle (1+x)^{m+1} $ :

$\displaystyle 1 + (m+1)x + \frac{m(m+1)x^2}{2} $

2) Used the binomial theorum to expand $\displaystyle (1 - 2x)^m $ :

$\displaystyle 1 - 2mx + {2m(m-1)x^2} $

3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?