Has anyone heard of this maths formula and where can I find the proof to check my proof is correct? $\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2$ The formula basically is:
The sum of all integers before and including $n$, plus all the integers up to and including $n-1$.
This will find $n^2$.
$$
\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2
$$
 A: $$\begin{array}{ccccccc}&&&\square&&&\\
&&\blacksquare&\square&\square\\
&\blacksquare&\blacksquare&\square&\square&\square\\
\blacksquare&\blacksquare&\blacksquare&\square&\square&\square&\square
\end{array}
\left.\rightarrow\quad
\begin{array}{cccc}
\square&\blacksquare&\blacksquare&\blacksquare\\
\square&\square&\blacksquare&\blacksquare\\
\square&\square&\square&\blacksquare\\
\square&\square&\square&\square
\end{array}\quad\right\}n\\
$$

In numbers,
$$\underbrace{\begin{array}{lrrrrrrrrr}
&n&+&n-1&+&n-2&+&\cdots&+&1\\
+&0&+&1&+&2&+&\cdots&+&n-1\\
\hline
&n&+&n&+&n&+&\cdots&+&n
\end{array}}_n$$

In summation signs,
$$\begin{align*}
\sum_{i=1}^ni + \sum_{i=1}^{n-1}i
&= \sum_{i=1}^ni + \sum_{i=0}^{n-1}i\\
&= \sum_{i=1}^ni + \sum_{j=1}^{n}(n-j) & (j = n-i)\\
&= \sum_{i=1}^n(i+n-i)\\
&= \sum_{i=1}^n n\\
&= n^2
\end{align*}$$
A: It is known that 
$$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$
Thus the value of your sum would be
$$\sum_{k=1}^nk+\sum_{k=1}^{n-1}k=\frac{n(n+1)}{2}+\frac{(n-1)(n)}{2}=\frac{n^2+n+n^2-n}{2}=\frac{2n^2}{2}=n^2.$$
A: This is equivalent to the well-known fact that the sum of the first $n$ odd numbers is $n^2$. For example, $1+3+5+7+9+11=36$. Why are they equivalent? Because of this:
\begin{align}
1+2+3+4+5+\phantom16&\\
{}+1+2+3+4+\phantom15&\\
-----------&\\
1+3+5+7+9+11&
\end{align}
A: Assuming you consider
$$
\sum^n_{i = 1}i = \frac{n(n+1)}{2}
$$
to be a well-known fact, observe that your sum is just
$$
\begin{array}{rcl}
\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i & = & \sum^n_{i=1}i + \sum^n_{i=1}i - n \\
&=& 2\sum^n_{i=1}i - n\\
&=& 2\frac{n(n+1)}{2} - n\\
&=& n(n+1) - n \\
&=& n^2 + n - n \\
&=& n^2
\end{array}
$$
A: In Zeilberger fashion: Plug in $n=2, 3$ into the LHS to get $4, 9$. Fit a quadratic to that and get $n^2$. Then to complete the proof, simply note
$$
\left(\sum^{n+1}_{i = 1}i + \sum^{n}_{i=1}i\right) - \left(\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i\right) = n+n+1 = (n+1)^2-n^2
$$
A: Note that $$\begin{align}i^2-(i-1)^2&\color{lightgray}{=2i-1}\\&=i\qquad+(i-1)\quad\quad\end{align}$$
Summing from $i=1$ to $n$ and telescoping LHS gives
$$\begin{align}n^2\qquad\quad&=\sum_{i=1}^ni+\sum_{i=1}^n(i-1)\\
&=\sum_{i=1}^n i+\sum_{i=0}^{n-1}i\\
&=\sum_{i=1}^n i+\sum_{i=1}^{n-1}i\qquad\blacksquare\end{align}$$
A: Ah man. I can't believe I'm late to this party. I discovered this as well a lot of years ago and came up with my own set of proofs.
I noticed that:
$$1 + 2 + .. + (n -1) + n + (n - 1) + ... + 2 + 1 = n^2$$
(which is the same thing that you have)
Proof by Induction:
Base case:
For n = 1:
$LHS = 1 = 1^2 = RHS$
Assuming that it is true for an integer $k > 1$:
(i.e. $1 + 2 + ... + (k - 1) + k + (k -1 ) + ... + k = k^2$)
The case for k + 1 becomes:
$LHS = 1 + 2 + ... + k + (k + 1) + k + ... + 2 + 1$
$ = k^2 + (k + 1) + k$ (using the induction hypothesis)
$ = (k + 1)^2 = RHS $
A: Legend wants that Carl Friedrich Gauss discover the formula
$$
\sum_{i=1}^n i = \dfrac{n(n+1)}{2}
$$
when he was six. Not surprising, since gaussing, ehm, guessing "Gauss" when trying to remember who found a certain result has a non trivial probability of success...
A: $$ \frac{(n-1) n}{2}+\frac{ n(n+1)}{2} = n^2.$$
