Closed form of summation with n+1 on top? How do you find the closed form of summations such as:
$$\sum\limits_{k=1}^{n+1} k$$
 A: $$\displaystyle\sum\limits_{k=1}^{n+1} k = \left( \sum\limits_{k=1}^{n} k \right) + (n + 1)$$ because $n+1$ is the last term.
A: mathematical notation is like a language and its correct use depends upon knowledge of syntax and conventions. the expression:
$$
\sum_{k=1}^{n+1} f(k)
$$
means
$$
f(1)+f(2)+f(3)+\dots +f(n)+f(n+1)
$$
a similar expression
$$
\sum_{k=1}^{n+\frac12} f(k)
$$
is recognized as meaningless because of unstated conventions which govern the use of the symbols. look at the grammatical convention which makes the following a true statement:
$$
1 = \frac12 + \frac14 + \frac18+ \dots
$$
it would be inappropriate to conclude from this that
$$
\dots=\frac18
$$
since the same symbol $\dots$ has a meaning which is determined by the context.
if i have misunderstood what you are puzzled about, and you simply want to evaluate the particular expression, then note that if
$$
S_f(n) = \sum_{k=1}^n f(k)
$$
then necessarily (although this requires a little thought) you must also have
$$
S_f(n) = \sum_{k=1}^n f(n+1-k)
$$
from this follows, by addition:
$$
2S_f(n) = \sum_{k=1}^n [f(k)+f(n+1-k)]
$$
in the particularly simple case where $f(k)=k$ this reduces to
$$
2S_f(n) = \sum_{k=1}^n [k+n+1-k] = \sum_{k=1}^n [n+1]
$$
since $n+1$ is independent of $k$ then we are simply summing $n$ terms, each of which is $n+1$ so the sum must be $n(n+1)$, and for this particular $f$ we have:
$$
2S_f(n) = n(n+1)
$$
hence:
$$
S_f(n) = \frac12 n(n+1)
$$
which is the $n^{th}$ triangular number
