## Find the sum of squares $S = \sum_{k=1}^{n}k^2$ [Proof by Algebra]

$\sum_{k=1}^{n}((k+1)^3 - k^3) = \sum_{k=1}^{n}(k^2+1+2k)(k+1) - k^3$

$\sum_{k=1}^{n} (k^3+k+2k^2+k^2+1+2k)-k^3$

$\sum_{k=1}^{n} (k^3+3k^2+3k+1)-k^3$

$\sum_{k=1}^{n} (3k^2+3k+1)$

$\Rightarrow \sum_{k=1}^{n}(k+1)^3 - \sum_{k=1}^{n}k^3 =(n+1)^3 - 1$

$\Rightarrow 2^3 + 3^3 + ... + n^3 + (n+1)^3 - (1 + 2^3 + 3^3 + ... + n^3) = (n+1)^3 -1$

$\Rightarrow (n+1)^3-1 = \sum_{k=1}^{n}(3k^2+3k+1)$

$\Rightarrow (n+1)^3-1 = 3\sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n} k + n$

$\Rightarrow (n+1)^3-1 = 3\sum_{k=}^{n} k^2 + (n+1) n \frac{3}{2} + n$

$\Rightarrow (n+1)^3-1 - (n+1) n \frac{3}{2} - n= 3\sum_{k=1}^{n} k^2$

$\Rightarrow (n+1)((n+1)^2-n \frac{3}{2})-(n+1) = 3\sum_{k=1}^{n} k^2$

$\Rightarrow (n+1)( (n+1)^2 -n\frac{3}{2}-1) = 3\sum_{k=1}^{n} k^2$

$\Rightarrow (n+1)( n^2+1+2n-n\frac{3}{2} - 1) = 3\sum_{k=1}^{n} k^2$

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

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

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

## Prove $S = 1 + 3 + ... + (2n-1) = n^2$ [ Proof by Picture ]

### From the picture, each number represents each colors(row/column)

$1 + 3 + 5 + 7 + 9 + 11 = 6^2$

$\Rightarrow 1 + 3 + 5 + (2n-1) = n^2$

## Prove $S = 1 + 3 + ... + (2n-1) = n^2$

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

$\Rightarrow 1 + 3 + 5 + ... + (2n-1) = 4(2n)^2 = n^2$