Proof for sum of consecutive number from one to any odd number is $(k+1)(2k+1)$ I recently saw a formula that was found by Levi Ben Gershon in 13th century.
The formula allows you to sum up the integers from one to any odd number by multiplying the middle element with the last.
$$ \sum_{n=1}^{2k+1}n = (k+1)(2k+1)$$
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Is there any intuitive way to explain why the formula works?
How can this be proved?
 A: So you want to calculate $1+2+3+\cdots+2k+1$.
Firstly, how many numbers are there? Clearly there are $n = 2k+1$ of them.
Secondly, what's the mean $\mu$? Because there are an odd number of integers being added, the integer in the middle of your set is precisely $\mu$. Thus $\mu = k+1$.
Now you also know that $\mu = \frac{1+2+3+\cdots+2k+1}{n}$ by the definition of a mean. By rearranging this, $(k+1)(2k+1) = \mu n = 1+2+3+\cdots+2k+1$
A: The sum of $N$ terms in arithmetic progression is $N$ times the average between the first and last one (which does in fact follow directly from the linearity of an arithmetic progression).
In this case there are $N=2k+1$ terms in an arithmetic progression with common difference $1\,$, and the average between the extreme terms is $\frac{1}{2}\left(1+(2k+1)\right) = k+1\,$.
A: A more fundamental formula is the one for the triangular numbers, which is the sum of all the numbers from $1$ to $n$ is $\frac 12n(n+1)$.  The intuitive way to see it is to add
$$1+2+3+\ldots +n\\\underline{n+(n-1)+(n-2)+\ldots 1}\\(n+1)+(n+1)+(n+1)+\ldots (n+1)$$ and in the sum there are $n$ terms of $n+1$, so $n(n+1)$ is twice the sum of the numbers from $1$ to $n$.  Once you have that you can derive the relation you cite.
