how do we find the sum of n terms of series : (1^2) + (1^2 +2^2) + (1^2 +2^2+3^2)+(1^2 +2^2+3^2+4^2) + ......
(not a multiple choice question)
Answer[spoiler](n*((n+1)^2)*(n+2)) /12[/spoiler]
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another one on sequence
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earth@work
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earth@work
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Got the solution from another forum, pasting it 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.
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.
