# Project Euler

## Sum square difference

### Problem 6

The sum of the squares of the first ten natural numbers is,

12 + 22 + … + 102 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + … + 10)2 = 552 = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Finally, something looks familiar. Remember Problem 1 and the summation formula. Instead of O(n) complexity, simply apply the summation formula to get O(1) complexity for this problem as follow:

$1^2 + 2^2+ \ldots + 100^2 - (1 + 2 + \ldots + 100)^2 = \sum_{i=1}^{100} i^2 - (\sum_{i=1}^{100} i)^2$

The summation of $i$ was discussed previously in Problem 1 as:

$\sum_{i=1}^n i = \frac{n(n+1)}{2}$

The summation of $i^2$ is:

$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$