# Playing with squares

Extending from particular examples I've found that $$n^2=\sum_{i=1}^{i=n-1} 2\, i+n$$ this is that for any square of side $$n$$ the area can be calculated in a simple way.

Example

For a square of side $$7$$, the result is: $$2×1+2×2+2×3+\cdots + 7=49$$

Question

Is there any way to prove this generally true, is there more than one way?Can you show at least one?

A visual proof. A variant of the one found in the book "Proofs Without Words"

Permuting your equation, i.e. moving $$n$$ to the left, and dividing by $$2$$, yields the equivalent equation

$$\frac{n(n-1)}{2}=\sum_{i=1}^{n-1} i$$ which is a very well known equation that can easily be proven by induction. Or just look at how Gauss did it when he was a child.

Note that the sum can be written as a telescopic sum: \begin{align}\sum_{i=1}^{n-1} 2 i+n&=\sum_{i=1}^{n-1} (2 i+1)+1=\sum_{i=1}^{n-1} ((i+1)^2-i^2)+1\\ &=(n^2-1)+1=n^2. \end{align} Similarly, we can show $$\sum_{i=1}^{n-1} (3i^2+3i)+n=n^3\quad\text{or}\quad \sum_{i=1}^{n-1} (4i^3+6i^2+4i)+n=n^4.$$

• wow, it is just impressive. there is something strange in the first equality, though... – santimirandarp May 21 at 11:40
• @santimirandarp Yes, you are right. I fixed it. – Robert Z May 21 at 11:48
• Nice! Now the last number appears correctly. But I found the (2i + 1) difficult notation, do you? Maybe it's just a matter of habit... – santimirandarp May 21 at 11:50
• $\sum_{i=1}^{n-1}1$ means the sum of $n-1$ times the number $1$, that is $n-1$. – Robert Z May 21 at 12:00
• Yes, so the mistake is supposing it needs an argument like $i$. Don't know where it comes from. Thanks again. – santimirandarp May 21 at 12:02

$$\sum_{i=1}^{i=n-1} 2\, i\,+n=2\left(\dfrac {n (n-1)}2\right)+n=n^2$$