Find a formula for the sum of the first $n$ even positive numbers

Question

I'm having a hard time grasping the concept of creating a formula to represent: $0 + 2 + 4 + ... + 2n$ when $n$ is a nonnegative integer.

Looking it up I see that it is $n(n+1)$

However I can't figure out how to get there, I'm lost on the steps that go from $\sum_{i=0}^n 2i $ to $ = n(n+1)$

I tried solving $a_nr^n+a_{n-1}r^{n-1}+a_{n-2}r^{n-2} = 0$ for $a_0 = 0$, $a_1=2$, $a_2=4$... but I don't think thats correct.

Answer 1

HINT

First $n$ even numbers look like $$ 2 + 4 + 6 + \ldots (2n-4) + (2n-2) + 2n. $$ Sum them pairwise, grouping first with last, 2nd with next-to-last, etc, and you will find each sum is exactly $2n+2$.

How many pairs are there, and are there any non-paired elements?

Answer 2

Hint:

$$\sum_{i=0}^n 2i =\sum_{j=0}^n 2(n-j)=2n(n+1)-\sum_{j=0}^n 2j$$

Answer 3

Note that\begin{align}2\times\bigl(2+4+\cdots(2n)\bigr)&=2+4+\cdots(2n)+(2n)+(2n-2)+\cdots+2\\&=\bigl(2+(2n)\bigr)+\bigl(4+(2n-2)\bigr)+\cdots+\bigl((2n)+2\bigr)\\&=n\times(2n+2).\end{align}