Has anyone heard of this maths formula and where can I find the proof to check my proof is correct? $\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2$

Question

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 $n^2$.

$$ \sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2 $$

Answer 1

It is known that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$ Thus the value of your sum would be $$\sum_{k=1}^nk+\sum_{k=1}^{n-1}k=\frac{n(n+1)}{2}+\frac{(n-1)(n)}{2}=\frac{n^2+n+n^2-n}{2}=\frac{2n^2}{2}=n^2.$$

Answer 2

$$\begin{array}{ccccccc}&&&\square&&&\\ &&\blacksquare&\square&\square\\ &\blacksquare&\blacksquare&\square&\square&\square\\ \blacksquare&\blacksquare&\blacksquare&\square&\square&\square&\square \end{array} \left.\rightarrow\quad \begin{array}{cccc} \square&\blacksquare&\blacksquare&\blacksquare\\ \square&\square&\blacksquare&\blacksquare\\ \square&\square&\square&\blacksquare\\ \square&\square&\square&\square \end{array}\quad\right\}n\\ $$

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In numbers,

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

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In summation signs,

$$\begin{align*} \sum_{i=1}^ni + \sum_{i=1}^{n-1}i &= \sum_{i=1}^ni + \sum_{i=0}^{n-1}i\\ &= \sum_{i=1}^ni + \sum_{j=1}^{n}(n-j) & (j = n-i)\\ &= \sum_{i=1}^n(i+n-i)\\ &= \sum_{i=1}^n n\\ &= n^2 \end{align*}$$

Answer 3

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}

Answer 4

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

$$ \begin{array}{rcl} \sum^n_{i = 1}i + \sum^{n-1}_{i=1}i & = & \sum^n_{i=1}i + \sum^n_{i=1}i - n \\ &=& 2\sum^n_{i=1}i - n\\ &=& 2\frac{n(n+1)}{2} - n\\ &=& n(n+1) - n \\ &=& n^2 + n - n \\ &=& n^2 \end{array} $$

Answer 5

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...

Answer 6

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

$$ \left(\sum^{n+1}_{i = 1}i + \sum^{n}_{i=1}i\right) - \left(\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i\right) = n+n+1 = (n+1)^2-n^2 $$

Answer 7

$$ \frac{(n-1) n}{2}+\frac{ n(n+1)}{2} = n^2.$$

Answer 8

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 $$\begin{align}n^2\qquad\quad&=\sum_{i=1}^ni+\sum_{i=1}^n(i-1)\\ &=\sum_{i=1}^n i+\sum_{i=0}^{n-1}i\\ &=\sum_{i=1}^n i+\sum_{i=1}^{n-1}i\qquad\blacksquare\end{align}$$

Answer 9

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 $