Simplify sum using a combinatorial identity I want to show that
$$
\sum_{i=0}^n {n\choose i} \left(x^{i+1}y^{n-i} + x^i y^{n-i+1}\right) = \sum_{i=0}^{n+1}{n+1 \choose i}x^i y^{n+1-i}.
$$
I was thinking to split the sum on the LHR into 2 sums and then perform an index transformation on the first sum, so I would be able to use Pascal's rule. Unfortunately I wasn't able to find the right transformation so far, as I always end up with something like $\sum_{i=0}^{n+1}{n\choose i-1}$ for the first sum, which doesn't make sense for the binomial coefficient (because $i-1=-1$).
Any hints?
 A: $$ \sum_{i=0}^n {n\choose i} \left(x^{i+1}y^{n-i} + x^i y^{n-i+1}\right) $$
$$= x\sum_{i=0}^n {n\choose i} \left(x^{i}y^{n-i}\right)+y\sum_{i=0}^n {n\choose i} \left(x^{i}y^{n-i}\right)$$
$$=x(x+y)^n+y(x+y)^n=(x+y)^n(x+y)=(x+y)^{n+1}$$
$$
=\sum_{i=0}^{n+1}{n+1 \choose i}x^i y^{n+1-i}.
$$
A: If you apply the recurrence on binomial coefficients, you have two choices:
You think of $\dbinom n {-1}$ as undefined and you have to take great care not to apply the recurrence to $\dbinom n 0$ (which you tried to do), or you just define $\dbinom n {-1} = 0$ and then you are allowed to always apply the recurrence (which is much better for calculation).
(Note that it isn't necessary to apply the recurrence for this particular exercice, but it is absolutely allowed.)
A: May be you can calculate as following:
$$\begin{align*}
\sum_{i=0}^{n}{n\choose i}\left(x^{i+1}y^{n-i}+x^i y^{n-i+1}\right)
&=\sum_{i=0}^{n}{n\choose i}x^{i+1}y^{n-i}+\sum_{i=0}^{n}{n\choose i}x^i y^{n-i+1}\\
&=\sum_{i=1}^{n+1}{n\choose i-1}x^{i}y^{n-i+1}+\sum_{i=0}^{n}{n\choose i}x^i y^{n-i+1}\\
&=\sum_{i=0}^{n+1}{n\choose i-1}x^{i}y^{n-i+1}+\sum_{i=0}^{n+1}{n\choose i}x^i y^{n-i+1}\\
&=\sum_{i=0}^{n+1}\left[{n\choose i-1}+{n\choose i}\right]x^{i}y^{n-i+1}\\
&=\sum_{i=0}^{n+1}{n+1\choose i}x^i y^{n+1-i}
\end{align*}$$
