Evaluate the sum $1+2+3+...+n$

Question

How do we evaluate the sum:

\begin{equation*} 1+2+...+n \end{equation*}

I don't need the proof with the mathematical induction, but the technique to evaluate this series.

Answer 1

Turn it backwards and sum those two together. Notice something?

Answer 2

Hint:

$$\begin{align}2\cdot (1 + 2 + \cdots + n) &= 1 + 2 + \cdots + (n - 1) + n + \\& + n + (n-1) + \cdots + 2 + 1 \end{align}$$

Now, add the numbers in the same column together.

Answer 3

You should notice that if you add the first term with the last the sum of these two is $n+1$. Now add the second term with the prelast then you get that the sum of these two is again $n+1$. Continue with the same way.

Only one term will remain unattached. Can you extract the result that:

$$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$

This is what Gauss did when he evaluated for the first time this sum at school!!

Answer 4

The average term is $\frac {n+1}2$ and there are $n$ of them. To see this, take the average of $n$ and $1$, of $n-1$ and $2$ etc - at each stage the average of the pair is $\frac {n+1}2$ and if you have a single number left at the end, it will be the central one i.e. $\frac {n+1}2$.

This is not so slick as the reverse and add method, and is wholly equivalent to it. So this answer is just in case it helps you to understand. I have found the concept of average term $\times$ number of terms useful as a shorthand for arithmetic progressions in general, rather than remembering a formula.