Evaluate that Product edited!
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pending
Evaluate
The answer is 420.
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1 solution
k = 1 ∑ 5 k ( k + 1 ) ( k + 2 ) 1 ( 1 + 1 ) ( 1 + 2 ) + 2 ( 2 + 1 ) ( 2 + 2 ) + 3 ( 3 + 1 ) ( 3 + 2 ) + 4 ( 4 + 1 ) ( 4 + 2 ) + 5 ( 5 + 1 ) ( 5 + 2 ) 1 ( 2 ) ( 3 ) + 2 ( 3 ) ( 4 ) + 3 ( 4 ) ( 5 ) + 4 ( 5 ) ( 6 ) + 5 ( 6 ) ( 7 ) 6 + 2 4 + 6 0 + 1 2 0 + 2 1 0 = 3 0 + 6 0 + 3 3 0 = 9 0 + 3 3 0 = 4 2 0