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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]
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 15 Aug 2011, 02:39
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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of n and S(n), S(2n)=?
(A) 2*S(n)
(B) n*S(n)
(C) 2n*S(n)
(D) 2S(n)+n^2
(E) S(n)+2n^2
Pls help with the easiest explanation possible..thnx
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Re: Sequence is making me go bonkers!! [#permalink]
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15 Aug 2011, 03:22
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DeeptiM wrote: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of n and S(n), S(2n)=?
(A) 2*S(n)
(B) n*S(n)
(C) 2n*S(n)
(D) 2S(n)+n^2
(E) S(n)+2n^2
Pls help with the easiest explanation possible..thnx for starters u could use the substitution technique where n =2 ==> sn = 3 then s(2n) = s(4) = 10 only D satisfies however if ur looking to solve it mathematically, Sn = n(a1 + an)/2 since this is an AP with difference = 1 and starting term a = 1 we can rewrite an as a+ (n-1)d = 1 + (n-1) Therefore Sn = n(1+ n)/2 or n+n^2 = 2Sn -- (1) S(2n) similarly = 2n[1 + 2n] / 2 = n + 2n^2 = n + n^2 + n^2 we know from (1) S(2n) = 2Sn + n^2 hence answer D
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Re: Sequence is making me go bonkers!! [#permalink]
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15 Aug 2011, 04:27
"for starters u could use the substitution technique where n =2 ==> sn = 3
then s(2n) = s(4) = 10 only D satisfies"
Can you explain how you would get s(n) = 3 if n is 2. disregarding the format of the sequence, if n is 2, the sum of the sequence should be at least 12 (10...+ 2).
its clearly much quicker than doing it mathematically! but I did go the math route, and my only falter compared to your calculation is that I cannot see how you've got rid of the division by 2 in the S(n) calculations.
"Therefore Sn = n(1+ n)/2 or n+n^2 = 2Sn -- (1)"
Shouldn't n(1 + n) / 2 become n + n^2 / 2?
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Re: Sequence is making me go bonkers!! [#permalink]
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15 Aug 2011, 04:46
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DeeptiM wrote: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of n and S(n), S(2n)=?
(A) 2*S(n)
(B) n*S(n)
(C) 2n*S(n)
(D) 2S(n)+n^2
(E) S(n)+2n^2
Pls help with the easiest explanation possible..thnx Let's see the pattern: For n=5, the sequence will be {1,2,3,4,5} \(S(n)=S(5)=1+2+3+4+5\) 2n=2*5=10, the sequence will be {1,2,3,4,5,6,7,8,9,10} \(S(2n)=S(10)=1+2+3+4+5+6+7+8+9+10=1+2+3+4+5+(1+5)+(2+5)+(3+5)+(4+5)+(5+5)\) \((1+2+3+4+5)+(1+2+3+4+5)+(5+5+5+5+5)\) \(S(5)+S(5)+5*5=S(5)+S(5)+5^2=2S(5)+5^2\) Since, n=5 \(2S(5)+5^2=2S(n)+n^2\) In general terms, \(S(n)=1+2+3+4,...+n\) \(S(2n)=1+2+3+4,...+n+(1+n)+(2+n)+(3+n)+(4+n),...+(n+n)\) \(S(2n)=(1+2+3+4+...+n)+(1+2+3+4+...n)+(n+n+...n-times)\) \(S(2n)=S(n)+S(n)+n^2\) \(S(2n)=2S(n)+n^2\) Ans: "D"
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Re: Sequence is making me go bonkers!! [#permalink]
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15 Aug 2011, 04:59
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meshell wrote: "for starters u could use the substitution technique where n =2 ==> sn = 3
then s(2n) = s(4) = 10 only D satisfies"
Can you explain how you would get s(n) = 3 if n is 2. disregarding the format of the sequence, if n is 2, the sum of the sequence should be at least 12 (10...+ 2).
its clearly much quicker than doing it mathematically! but I did go the math route, and my only falter compared to your calculation is that I cannot see how you've got rid of the division by 2 in the S(n) calculations.
"Therefore Sn = n(1+ n)/2 or n+n^2 = 2Sn -- (1)"
Shouldn't n(1 + n) / 2 become n + n^2 / 2? Michelle, the series is 1,2,3,4,.... and Sn is the sum of the series until n terms .. so the sum of the series for 2 terms or s(2) = 1+2 = 3 and s(4) = 1+2+3+4 = 10 i hope this helps explain your concern on "disregarding the format of the sequence, if n is 2, the sum of the sequence should be at least 12 (10...+ 2). " if you still have questions, i'll be happy to help. on the mathematical formula yes sn = [n(1+n)] / 2 and is therefore indeed sn = [n + n^2] / 2 but to avoid confusion, i have pulled the 2 to the other side making it 2* Sn = [n + n^2] so (n + n^2 ) equals 2*Sn and not just Sn.
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Re: Sequence is making me go bonkers!! [#permalink]
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15 Aug 2011, 05:02
fluke wrote: DeeptiM wrote: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of n and S(n), S(2n)=?
(A) 2*S(n)
(B) n*S(n)
(C) 2n*S(n)
(D) 2S(n)+n^2
(E) S(n)+2n^2
Pls help with the easiest explanation possible..thnx Let's see the pattern: For n=5, the sequence will be {1,2,3,4,5} \(S(n)=S(5)=1+2+3+4+5\) 2n=2*5=10, the sequence will be {1,2,3,4,5,6,7,8,9,10} \(S(2n)=S(10)=1+2+3+4+5+6+7+8+9+10=1+2+3+4+5+(1+5)+(2+5)+(3+5)+(4+5)+(5+5)\) \((1+2+3+4+5)+(1+2+3+4+5)+(5+5+5+5+5)\) \(S(5)+S(5)+5*5=S(5)+S(5)+5^2=2S(5)+5^2\) Since, n=5 \(2S(5)+5^2=2S(n)+n^2\) In general terms, \(S(n)=1+2+3+4,...+n\) \(S(2n)=1+2+3+4,...+n+(1+n)+(2+n)+(3+n)+(4+n),...+(n+n)\) \(S(2n)=(1+2+3+4+...+n)+(1+2+3+4+...n)+(n+n+...n-times)\) \(S(2n)=S(n)+S(n)+n^2\) \(S(2n)=2S(n)+n^2\) Ans: "D" Thanks Fluke for saving my back on so many occasions  kudos to u!!
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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]
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08 Sep 2014, 00:21
W.K.T \(S(n)=\frac{n(n+1)}{2}\) ---- first relation \(S(2n)=\frac{2n(2n+1)}{2}\) \(S(2n)=2n(\frac{n}{2}+\frac{n+1}{2})\) \(Substitute \frac{n+1}{2} = \frac{S(n)}{n} --from-1st-relation\) \(S(2n)=2n(\frac{n}{2}+\frac{S(n)}{n})\) reduce \(S(2n) = n^2 + 2S(n)\) Ans : D
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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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08 Sep 2014, 00:21
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