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# how do we find the sum of n terms of series : (1^2) + (1^2

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Intern
Joined: 11 Oct 2008
Posts: 10
how do we find the sum of n terms of series : (1^2) + (1^2 [#permalink]

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Updated on: 30 Oct 2008, 09:35
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) + ......

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Originally posted by earthwork on 29 Oct 2008, 21:01.
Last edited by earthwork on 30 Oct 2008, 09:35, edited 2 times in total.
SVP
Joined: 29 Aug 2007
Posts: 2427

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29 Oct 2008, 22:15
earthwork wrote:
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) + ......

no clue..
Probably big shots like walker have the solution.
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Intern
Joined: 11 Oct 2008
Posts: 10

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29 Oct 2008, 22:40
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!
VP
Joined: 17 Jun 2008
Posts: 1479

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29 Oct 2008, 22:52
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.

I may be wrong.
Current Student
Joined: 28 Dec 2004
Posts: 3292
Location: New York City
Schools: Wharton'11 HBS'12

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29 Oct 2008, 23:03
it cant be progressive, I think geometric series is more like it..
Intern
Joined: 30 Oct 2008
Posts: 30

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30 Oct 2008, 06:58
The sequence can be reduced to
(n*1) + (n-1)2^2 + (n-2)3^2 + (n-3)4^2.....
Intern
Joined: 11 Oct 2008
Posts: 10

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Updated on: 30 Oct 2008, 09:46
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.

Originally posted by earthwork on 30 Oct 2008, 09:31.
Last edited by earthwork on 30 Oct 2008, 09:46, edited 2 times in total.
SVP
Joined: 29 Aug 2007
Posts: 2427

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30 Oct 2008, 09:37
earthwork wrote:
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]
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Intern
Joined: 11 Oct 2008
Posts: 10

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30 Oct 2008, 09:45
this is a formula for sum of n terms for series:
∑k=n(n+1)/2
∑k^2=n(n+1)2n+1)/6
∑k^3=(n(n+1))^2/4

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: sequence   [#permalink] 30 Oct 2008, 09:45
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# how do we find the sum of n terms of series : (1^2) + (1^2

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