Sum - can I just change the terms? I am looking at an exercise where this sum appears $\sum_{k=0}^{n}\binom{2n+1}{2n+1-k}$, and I saw in my textbook that it should be equal to $\sum_{u=n+1}^{2n+1}\binom{2n+1}{u}$.
I replaced at the first sum $2n+1-k$ with $u$, and I get this sum: $\sum_{u=2n+1}^{n+1}\binom{2n+1}{u}$. How did they find the result? Can I just change the terms $n+1$ and $2n+1$, because $n+1$ is smaller?

It sometimes helps to write out the terms.
\begin{align*}
\sum_{k=0}^n\binom{2n+1}{2n+1-k} &= \binom{2n+1}{2n+1} + \binom{2n+1}{2n} + \dots + \binom{2n+1}{n+2} + \binom{2n+1}{n+1}\\
\\
\sum_{u=n+1}^{2n+1}\binom{2n+1}{u} &= \binom{2n+1}{n+1} + \binom{2n+1}{n+2} + \dots + \binom{2n+1}{2n} + \binom{2n+1}{2n+1}.
\end{align*}
So the two sums are the same, they just reverse the order of the summands. 
To see this via a change of index, let $u = 2n+1-(n-k)$ so that $u = n+1+k$; as $k$ goes from $0$ to $n$, $u$ goes from $n+1+0 = n+1$ to $n+1+n = 2n+1$. You can think of this as a combination of two index changes, first $w = n - k$ reverses the order of the terms, and then $u = 2n+1-w$ shifts the index. 
