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Intern
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how do we find the sum of n terms of series : (1^2) + (1^2 [#permalink]
 29 Oct 2008, 21:01
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) + ......
Last edited by earthwork on 30 Oct 2008, 09:35, edited 2 times in total.
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CEO
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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
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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!
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SVP
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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.
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Current Student
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it cant be progressive, I think geometric series is more like it..
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Intern
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The sequence can be reduced to
(n*1) + (n-1)2^2 + (n-2)3^2 + (n-3)4^2.....
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Intern
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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, 09:46, edited 2 times in total.
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CEO
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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
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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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