Sums of binomials with increasing powers

Question

I am curious if there is a way to simplify the following:

\begin{align} & 1+a+(a+b)^2+(a+b)^3+\cdots+(a+b)^n \\[8pt] = {} &1+a+\sum_{k=0}^2 {2 \choose k} a^kb^{2-k} +\sum_{k=0}^3 {3 \choose k}a^kb^{3-k}+\cdots \end{align}

Answer 1

Let $$S=1+(a+b)+(a+b)^2 + \dots + (a+b)^n$$ and the required is $S-b$, now $$S(a+b)=(a+b)+(a+b)^2 + \dots + (a+b)^n+(a+b)^{n+1}$$ $$S(a+b)-S=(a+b)^{n+1}-1 \iff S=\frac{(a+b)^{n+1}-1}{a+b-1}$$ So the required is $$\frac{(a+b)^{n+1}-1}{a+b-1}-b$$