# 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 $\binom{n}{k}$?

• 1) How did you reach/(where did you find) this? How we should understand $\binom{n}{n+1}$ is probably understood from the context - and perhaps it should be omitted. 2) The internal parameter $k$ runs from 1 to $n+1$ , hence we can replace $k$ with a new term (which we also call $k$) that runs from 0 to $n$. Commented Oct 13, 2016 at 8:50
• It is part of an induction proof of $\sum_{k=1}^{n}\binom{n}{k}ka^{k}b^{n-k}=na(a+b)^{n-1}$. My suspicion, confirmed by Yves' suggestion, was that we could set it to $0$ by convention but I wasn't quite sure. Commented Oct 13, 2016 at 9:33
• Thanks, perhaps you can prove that using analysis too..? Commented Oct 13, 2016 at 12:42

Since $\binom{n}{n+1}=0$, we have

$$\sum_{k=1}^{n+1}\binom{n}kka^kb^{n+1-k}=\sum_{k=1}^n\binom{n}kka^kb^{n+1-k}=b\sum_{k=1}^n\binom{n}kka^kb^{n-k}\;;$$

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 $k$.

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.

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}