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= 8k³ + 12k² -2k -3

Sn = summation of Kth term from k = 1 to k = n

Sn = 8{n(n+1)/2}² + 12n(n+1)(2n+1)/6 - 2n(n+1)/2 - 3n

Sn = 2{n(n+1)}² + 2n(n + 1)(2n+1) - n(n+1) -3n now you can arrange them according to answer.

re-check for any calculation error. if any doubt please ask.