again I'm stuck already on the first steps of an inductive proof of
$$ (a+b)^{n+1} = \sum_{k=0}^{n+1} {n+1\choose k}a^kb^{n+1-k} $$
that is, I'm trying to understand the solution to this. It starts with:
$$ (a+b)^{n+1} = (a+b) * (a+b)^n = (a+b) * \sum_{k=0}^{n} {n\choose k}a^kb^{n-k} $$
This part I understand, but then I get stuck with this next part (continuing the row above)
$$ = \sum_{k=0}^{n} {n\choose k}a^{k+1}b^{n-k}+\sum_{k=0}^{n} {n\choose k}a^kb^{n+1-k} $$
Don't understand where the first term comes from and why in the second it's now $ b^{n+1-k} $. I'm a bit of a noob I have to say, so don't be so hard on me ;)
the rest of the proof I don't understand either, but one step at a time :).
Thanks!!
$$\begin{align}(a+b)\sum_{k=0}^{n} {n\choose k}a^kb^{n-k} &= a\sum_{k=0}^{n} {n\choose k}a^kb^{n-k}+b\sum_{k=0}^{n} {n\choose k}a^kb^{n-k}\\ &= \sum_{k=0}^{n} {n\choose k}a^{k+1}b^{n-k}+\sum_{k=0}^{n} {n\choose k}a^kb^{n+1-k}\end{align}$$
First, the $\Sigma$ sum was distributed over the sum $a+b,$ then $a$ and $b$ were distributed over the $\Sigma$ sums.