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i cud only reach till (1^2) + (1^2 +2^2) + (1^2 +2^2+3^2)+(1^2 +2^2+3^2+4^2) + ...... = (n*1^2)+((n-1)*2^2)+((n-2)*3^2)+...... dnt even know if this is correct step!

Is it not an arithmatic progression where each subsequent term is more than the previous term by n^2 and hence, d = n^2, a = 1^2 and the sum should be a + (n-1)d = 1 + (n-1)*n^2.

Finally someone did help me solve it in another forum, solution below : SOLUTION : nth term = n(n+1)(2n+1)/6 = (2n^3 + 3n^2 + n)/6 Sum = 1/6{ 2.∑(n^3) + 3.∑(n^2) + ∑(n)} = (1/6)[{2.(n^2)(n+1)^2/4} + 3. n(n+1)(2n+1)/6 + n(n+1)/2] Now we can simplify this to get final answer.

Answer: (n*((n+1)^2)*(n+2)) /12

Last edited by earthwork on 30 Oct 2008, 08:46, edited 2 times in total.

Finally someone did help me solve it in another forum, solution below : SOLUTION : nth term = n(n+1)(2n+1)/6 = (2n^3 + 3n^2 + n)/6 Sum = 1/6{ 2.∑(n^3) + 3.∑(n^2) + ∑(n)} = (1/6)[{2.(n^2)(n+1)^2/4} + 3. n(n+1)(2n+1)/6 + n(n+1)/2] Now we can simplify this to get final answer.

can you explain this: = (1/6)[{2.(n^2)(n+1)^2/4} + 3. n(n+1)(2n+1)/6 + n(n+1)/2] _________________