Index Shifting & Simplifying Taylor Expansion I am doing a proof by induction for a taylor expansion $(1+a)^n = \sum_{k=0}^{n} {n \choose k} a^k$ and I have made some good progress, but not sure of shifting indices while combining summations. 
The base case is pretty obvious. In the inductive hypothesis, I set $n=m$ for $m \ge 2$, assume that $(1+a)^m = \sum_{k=0}^{m} {m \choose k} a^k$ and work my way through the expansion to show that it is true for $n=m+1$ and after a few steps I arrive at
$$(1+a)^1(1+a)^m = \sum_{k=0}^{m} {m \choose k} a^k + \sum_{k=0}^{m} {m \choose k} a^{k+1}$$
$$ = \sum_{k=0}^{m} {m \choose k} a^k + \sum_{k=1}^{m+1} {m \choose k-1} a^{k}$$
I know I can use pascals formula to combine the combinations inside the summations, but I don't know how to fix the index of the first summation term. Any help is appreciated .
Regards,
 A: Maybe it is easier to do the induction step the other way round and start with the sum. This way it could be easier to see how to split sums and shift indices.

Induction step: $m\rightarrow m+1$
We obtain
  \begin{align*}
\sum_{k=0}^{m+1}\binom{m+1}{k}a^k&=\binom{m+1}{0}a^0+\sum_{k=1}^m\binom{m+1}{k}a^k+\binom{m+1}{m+1}a^{m+1}\tag{1}\\
&=a^0+\sum_{k=1}^m\left(\binom{m}{k}+\binom{m}{k-1}\right)a^k+a^{m+1}\tag{2}\\
&=\sum_{k=0}^m\binom{m}{k}a^k+\sum_{k=0}^{m-1}\binom{m}{k}a^{k+1}+a^{m+1}\tag{3}\\
&=(1+a)^m+a\sum_{k=0}^{m}\binom{m}{k}a^{k}\tag{4}\\
&=(1+a)^m+a(1+a)^m\tag{5}\\
&=(1+a)^{m+1}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we separate the terms with $k=0$ and $k=m+1$ as preparation for the next step.

*In (2) we apply the binomial identity $\binom{m+1}{k}=\binom{m}{k}+\binom{m}{k-1}$.

*In (3) we merge $a_0$ into the left sum and shift the index in the right sum.

*In (4) we apply the induction hypothesis once and merge $a^{m+1}$ into the right sum.

*In (5) we apply the induction hypothesis once again.
