What is the number of different sums of the items of a set of consecutive natural numbers? I have been playing around with sets of consecutive natural numbers like say $S=\{1,2,3,...,10\}$ and I have come up with this problem for which however I have not yet found an answer and I don't know if there is one. My problem is the following: Considering a set $S=\{1,2,3,...,n\}$ where $n$ is an arbitrary natural number, can we derive a formula that will indicate the number of different sums that will be produced by summing the items of the set $S$ in all different combinations? For example for the set $S=\{1,2,3,4,5\}$ we get:
$$1+2=3\\
1+2+3=6\\
1+2+3+4=10\\1+3+4+5=13\\1+2+4+5=12\\1+2+3+4+5=15\\
1+2+3+5=11\\1+4+5=10\\1+3+5=9\\1+3+4=8\\1+2+5=8\\1+2+4=7\\1+3=4\\1+4=5\\1+5=6\\
2+3=5\\2+4=6\\2+5=7\\
2+3+4=9\\2+3+5=10\\2+4+5=11\\
2+3+4+5=14\\
3+4=7\\3+5=8\\
3+4+5=12\\
4+5=9\\$$
So we have $26$ different summations and $13$ different sums(I hope I didn't make any mistake)

Here is an answer with the same content as Thomas Andrews' answer, but presented a bit more heuristically.
Since the sum of the first $n$ positive integers is given by $\frac{n(n+1)}{2}$, you can obviously make that sum with the $n$ integers. By removing one integer addend at a time, you can make a number which is $1$ smaller, and $2$ smaller, etc. all the way up to $n$ smaller, which will provide sums equal to all of the integers down to $\frac{n(n+1)}{2}-n$
But $\frac{n(n+1)}{2}-n=\frac{n(n+1)}{2}-\frac{2n}{2}=\frac{n(n+1)-2n}{2}=\frac{n(n+1-2)}{2}=\frac{(n-1)n}{2}$
You should recognize the final expression as the form of the sum of the first $n-1$ positive integers, which in any event is what you have when you remove $n$ as the addend from the set of the first $n$ integers.
You can reductively repeat this process to generate all smaller integer sums until you reach the set $\{1,2\}$, which can give you the sum $3$, but there is no way (using sums of two or more integers) that you can generate the sum $2$ or $1$.
Hence, the set of the first $n$ positive integers can generate $\frac{n(n+1)}{2}-2$ distinct sums.
