How can I determine the function for summing a series of integers? I want to sum a series of integers, say from 1 to 5, but I want to generalise this function. I have seen that the function is:
$$
\text{sum} = \frac{n(n+1)}{2}
$$ 
where $n$ is the number to sum up to
So for $n=5$, i.e. $1 + 2 + 3 + 4 + 5$, the sum is $15$.
My question is, how could I (without knowing by heart) work out that function myself?
 A: I'll show by building from a few concepts.
The following technique is called telescoping,
$$\sum_{j=1}^{n}(a_j - a_{j-1}) = a_n - a_0$$
proof,
$$a_1 + a_2 + \dots + a_n +$$
$$(-a_0) + (-a_1) + \dots +(-a_{n-1})$$
Adding them all up, we end up with $a_n - a_0$.
Next I will show you the sum of first $n$ odd numbers. Note the identity
$$k^2 - (k-1)^2 = 2k-1, k\in \mathbb{Z}$$
where $2k-1$ is the $k^{th}$ odd number. The sum of the first $n$ odd number is then
$$\sum_{k=1}^{n}(2k-1)$$
Using the identity mentioned above,
$$\sum_{k=1}^{n}(2k-1)$$
$$=\sum_{k=1}^{n}(k^2 - (k-1)^2)$$
which by telescoping, is
$$1^2 + 2^2 + \dots + n^2 +\\
(-0^2) - 1^2 - 2^2 \dots -(n-1)^2\\
= n^2$$
The sum of the first $n$ odd numbers is then $n^2$. Finally we use all of these to prove the sum of first $n$ integers. Note that the $k^{th}$ odd number is,
$$2k-1$$
and the summation for first $n$ odd number is,
$$\sum_{k=1}^{n}(2k-1)$$
We want to get rid of $2k-1$ to just $k = (2k-1)-k+1$, thus the sum of first $n$ natural numbers is,
$$\sum_{k=1}^{n}k = \sum_{k=1}^{n}(2k-1 - k + 1)$$
$$\sum_{k=1}^{n}k = \sum_{k=1}^{n}(2k-1) - \sum_{k=1}^{n}k + \sum_{k=1}^{n}1$$
$$\sum_{k=1}^{n}k + \sum_{k=1}^{n}k = \sum_{k=1}^{n}(2k-1) + \sum_{k=1}^{n}1$$
We already know the sum of first $n$ odd and $\sum_{k=1}^{n}1 = n$, therefore
$$2\sum_{k=1}^{n}k = n^2 + n$$
$$\sum_{k=1}^{n}k = \dfrac{n^2 + n}{2}$$
$$\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}$$
This is just one of them, there are lots more. But for me this is the simplest.
A: Let $$S = 1 + 2 + 3 + ... + n$$
Rearrange to the form $$S = (n + 1) + ((n-1) + 2) + ... + (n - (n - 1) + n)
=(n+1)+(n+1)+...(n+1)$$
So there are $n/2$ terms (because we've paired them off), so $$S = \frac{n}{2}(n+1) = \frac{n(n+1)}{2}$$
A: Define $S(n)$ as:
$$S(n)=\sum_{k=1}^n k$$
Then it is easy to see that $S(n)$ is a second degree polynomial in $n$. Therefore
$$S(n)= \frac{1}{2}(n-2)(n-3)S(1) -(n-1)(n-3)S(2) + \frac{1}{2}(n-1)(n-2)S(3) = \frac{1}{2}n (n+1)$$
A: This formula is usually attributed to Gauss, who worked it out in a really simple way (the legend say, to finish earlier the boring homework of summing all naturals up to 100)
Think of numbers as sums of 1s.
$$1$$
$$1+1$$
$$1+1+1$$
$$1+1+1+1$$
$$\cdots$$
If you try to calculate the area of this triangle (where area = number of ones) you get, simply, Gauss' formula
$$A=sum=\frac{n(n+1)}{2}$$
Where the $n+1$ is justified by the fact that your upper vertex has not a lenght of $0$. You can imagine it as a trapeze with bases of length $1$ and $n$ and height of $n$.
