That the pattern continues forever can also be proven algebraically: ∑ ∑ k = 1 n ( 2 k − − 1 ) = 1 2 ( ∑ ∑ k = 1 n ( 2 k − − 1 ) + ∑ ∑ k = 1 n ( 2 ( n − − k + 1 ) − − 1 ) ) = 1 2 ∑ ∑ k = 1 n ( 2 ( n + 1 ) − − 1 − − 1 ) = n 2