Binomial theorem shifting numeration I am considering the following expression:
$$\sum_{k=1}^{n+1}\left[\binom{n}{k-1}+\binom{n}{k}\right]ka^{k}b^{n+1-k} = a\sum_{k=1}^{n+1}\binom{n}{k-1}ka^{k-1}b^{n-(k-1)}+b\sum_{k=1}^{n}\binom{n}{k}ka^{k}b^{n-k}$$
1) How can we drop $n+1$ in the second sum? The last binomial coefficient in it doesn't make sense then but how did we end up with such a non-existent term in the first place?
We proceed to get:
$$a\sum_{k=0}^{n}\binom{n}{k}(k+1)a^{k}b^{n-k} + b\sum_{k=1}^{n}\binom{n}{k}ka^{k}b^{n-k}$$
2) Now, how did we get the first sum? I get that we can shift the numeration by $1 $and $a, b, k$ give the same result but why can we write the shifted binomial coefficient as $\binom{n}{k}$?
 A: Since $\binom{n}{n+1}=0$, we have
$$\sum_{k=1}^{n+1}\binom{n}kka^kb^{n+1-k}=\sum_{k=1}^n\binom{n}kka^kb^{n+1-k}=b\sum_{k=1}^n\binom{n}kka^kb^{n-k}\;;$$
this answers your first question. 
For your second question, the change of $\binom{n}{k-1}$ to $\binom{n}k$ is just a matter of carrying out the index shift completely. Let $\ell=k-1$, so that $k=\ell+1$; then
$$\sum_{k=1}^{n+1}\binom{n}{k-1}ka^{k-1}b^{n-(k-1)}=\sum_{\ell=0}^n\binom{n}\ell(\ell+1)a^\ell b^{n-\ell}\;,$$
and we conclude by renaming $\ell$ back to $k$.
A: It is a natural convention to define $$\binom nk=0$$ for $k<0$ and $k>n$.
This is coherent with Pascal's Identity and with the other natural convention that $k!=\pm\infty$ for $k<0$.
Then you may write an "unbounded" binomial theorem
$$(a+b)^n=\sum_{k\in\mathbb Z}\binom nka^kb^{n-k}$$ which can be handy.
A: It is difficult to answer your first question, because I do not know how the LHS of your first equation was reached. For your second question, if you get confused with change in the summation limits, the most safe way to tackle it is to write down some terms. Here, your summation starts at $k=1$ and ends at $k=n+1$ and you want to start from $k=0$ and end at $k=n$ (for some reason). In this specific case, you can do this easily by observing that everywhere $k$ appears with a $-1$ next to it. So, actually you have $k-1$ with $k$ running from $1$ to $n+1$ which equivalently can be written as $k$ running from $0$ to $n$. But, for the general case, as mentioned you can write down the first few and last terms to see what is going on (I highlighted with blue all instances of $k$): 
\begin{align}\sum_{k=1}^{n+1}\binom{n}{k-1}ka^{k-1}b^{n-(k-1)}&=\underbrace{\dbinom{n}{\color{blue}0}\color{blue}1\cdot a^{\color{blue}0}\cdot b^{n-\color{blue}0}}_{k=1}+\underbrace{\dbinom{n}{\color{blue}1}\color{blue}2\cdot a^{\color{blue}1}\cdot b^{n-\color{blue}1}}_{k=2}+\\[0.2cm]&\hspace{40pt}\dots+\underbrace{\dbinom{n}{\color{blue}n}(\color{blue}{n+1})\cdot a^{\color{blue}n}\cdot b^{n-\color{blue}n}}_{k=n+1}\\[0.4cm]&=\sum_{k=0}^{n}\binom{n}{k}(k+1)a^{k}b^{n-k}\end{align}
