To clarify this proof, since the n odd positive integer is m : : = 2 n − − 1 , if S : : = ∑ ∑ k = 1 n ( 2 k − − 1 ) = 1 + 3 + ⋯ ⋯ + ( m − − 2 ) + m denotes the sum of the first n odd integers then S + S = 1 + 3 + ⋯ ⋯ + ( m − − 2 ) + m + m + ( m − − 2 ) + ⋯ ⋯ + 3 + 1 = ( m + 1 ) + ( m + 1 ) + ⋯ ⋯ + ( m + 1 ) + ( m + 1 ) ( n terms) = n ( m + 1 ) so that S = 1 2 n ( m + 1 ) . Substituting n = 1 2 ( m + 1 ) and m + 1 = 2 n gives, respectively, the formulas 1 + 3 + ⋯ ⋯ + m = 1 4 ( m + 1 ) 2 and 1 + 3 + ⋯ ⋯ + ( 2 n − − 1 ) = n 2 where the first formula expresses the sum entirely in terms of the odd integer m while the second expresses it entirely in terms of n , which is m 's ordinal position in the list of odd integers 1 , 3 , 5 , … … .