Sequence 1, 2, 3, 4, ...an...In terms of n, S(2n)-S(n)? I : PS Archive
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# Sequence 1, 2, 3, 4, ...an...In terms of n, S(2n)-S(n)? I

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Manager
Joined: 01 Nov 2007
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Sequence 1, 2, 3, 4, ...an...In terms of n, S(2n)-S(n)? I [#permalink]

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11 Mar 2008, 09:37
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Sequence 1, 2, 3, 4, ...an...In terms of n, S(2n)-S(n)?

I spent great amount of time to solve this problem, please advise the most effective approch to such kind of problems.
Manager
Joined: 20 Sep 2007
Posts: 106
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Kudos [?]: 69 [0], given: 0

Re: PS (Sequence) [#permalink]

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11 Mar 2008, 10:02
The best way to solve this is to use the formulae for sum of a series.which is
Sum of n terms= n/2[2a + (n-1)d]
where a= first term of series which is 1 here
n = no of terms which is n in first case and 2n in second case
d= difference between two terms
Hence
Sum of n terms = n/2[2+2n-1]=n/2[n+1]
Sum of 2n terms = 2n/2[2+2n-1]=n[2n+1]
Difference of both = n(3n+1)/2
Re: PS (Sequence)   [#permalink] 11 Mar 2008, 10:02
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# Sequence 1, 2, 3, 4, ...an...In terms of n, S(2n)-S(n)? I

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