Intuition to why average of the square of a positive integer and the integer itself is the sum of all numbers from 1 to the integer?

Question

The sum of all numbers from 1 to n, i.e.

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

This happens to be show that the average of a number and its square equals the sum of all numbers from 1 to that number.
Is there any intuitive explanation to this or does it merely end up being true?

Sorry if this is a duplicate; I couldn't find a similar question myself.

Answer 1

Well you can distribute n points on a circle and connect them with each other: For the first point you have n-1 lines to draw, for the second n-2 and so... But this also exactly the same as the amount of distinct pairs of points you can pick out of n points. $(n-1)n \over 2 $ = $ \binom{n}{2} $

Answer 2

An intuitive way to see why this formula works is to imagine all the numbers from $1$ to $n$ written in a line: $$1,2,3,4,5,...,n-3,n-2,n-1,n$$

Now sum the first with the last, the second with the second last and so on, the sum is always $n+1$. How many of these sums do we have? Half the length of the list, which contains $n$ numbers, that is $n\over 2$, so that the total is $\frac{n}{2}(n+1)$.

Answer 3

It is perhaps easier to see intuitively why $n^2+n$ is twice the $n$th triangle number. Divide by $2$ for your result. Here is a visualisation:

Answer 4

Maybe it is easier to see why the average $\frac{1}{n} \sum_{i=1}^n i $ is equal to the average $\frac{1 + n}{2}$?

Answer 5

Well, consider that it is quite clear that $$n\leq\sum_{i=1}^{n}i\leq n^2$$ where the first bound is $\sum_{i=1}^{n}1$, assuming that every term is $1$, the minimum term - and the upper bound is $\sum_{i=1}^{n}n$ - assuming that every term is $n$, the maximum term. When we take the average of these two bounds, we are more or less assuming that every element is equal to the average of the maximum and minimum - this works because the set of numbers $\{1,2,\ldots,n\}$ is a uniform distribution over some interval of integers, and hence its mean is simply the average of its minimum and maximum - and since everything in this problem is additive, the average of the bounds for the sum is equal to the sum of the average.