Answers

2014-06-08T23:08:22+05:30
Let Kth term = (2k-1)(2k+1)(2k+3)
                  = 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.

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