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find the sum of series

(4+6+8+...+2n-4)+(4+6+...+2n-6)+...+4

 Jul 29, 2016
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(4+6+8+...+2n-4)+(4+6+...+2n-6)+...+4 = (n^3 - 3n^2 - 4n + 12)/3.      n >= 4

 

 

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 Jul 29, 2016

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