$x^2$ in a different form?

Question

Is it known that $x^2=$ the sum of all of the numbers underneath it doubled, plus $x$?

Thought of it, thinking about pyramid push ups so $3^2= 2(1)+2(2)+3$,

$5^2= 2(1)+2(2)+2(3)+2(4)+5$

Answer 1

You are asking the following $$n^2 \overset{?}{=} 2(1 + 2 + \cdots + (n-1)) + n.$$

You have verified it for $n=3$. (You can check $n=2$ as well.)

For a proof by induction, you need to prove the inductive step. Specifically, if the above holds with $n=k$, can you prove it holds with $n=k+1$?

$$(k+1)^2 = k^2 + 2k + 1 = (2(1+\cdots+(k-1)) + k) + 2k + 1 = 2(1+\cdots+k) + (k+1) $$

Answer 2

Draw dots formed an $n\times n$ square, and split it to $3$ parts: one diagonal, dots below it, and dots above it.