Question

What is the sum of all odd numbers between 0 and 100?

Answer 1

First, notice an interesting pattern here:

`1, 4, 9, 16, 25, ...`

The differences between perfect squares (starting at `1-0 = 1`) is:

`1, 3, 5, 7, 9, ...`

The sum of `1+3+5+7+9` is `25`, the `5^"th"` nonzero square.

Let's take another example. You can quickly prove that:

`1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100`

There are `(19+1)/2 = 10` odd numbers here, and the sum is `10^2`.

Therefore, the sum of `1 + 3 + 5 + ... + 99` is simply:

`((99+1)/2)^2 = color(blue)(2500)`

Formally, you can write this as:

`color(green)(sum_(n=1)^N (2n-1) = 1 + 3 + 5 + ... + (2N - 1) = ((N+1)/2)^2)`

where `N` is the last number in the sequence and `n` is the index of each number in the sequence. So, the `50^"th"` number in the sequence is `2*50 - 1 = 99`, and the sum all the way up to that is `((99 + 1)/2)^2 = 2500`.