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Call the terms in the series a 1 , a 2 , … . Note the difference a n − a n − 1 is the n -th triangular number, and a 1 is the first triangular number. Therefore, when summing the first 5 0 0 terms, we are summing 5 0 1 − n copies of the n -th triangular number as n ranges from 1 to 5 0 0 . Then the answer is k = 1 ∑ 5 0 0 n = 1 ∑ k 2 n ( n + 1 ) = k = 1 ∑ 5 0 0 6 k ( k + 1 ) ( k + 2 ) = 2 4 5 0 0 ⋅ 5 0 1 ⋅ 5 0 2 ⋅ 5 0 3 ,
i.e. the answer is 2 6 3 5 5 3 1 3 7 5 .