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Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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15 Aug 2011, 03:27
"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.
Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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15 Aug 2011, 03:46
1
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, \(S(n)=1+2+3+4,...+n\) \(S(2n)=1+2+3+4,...+n+(1+n)+(2+n)+(3+n)+(4+n),...+(n+n)\) \(S(2n)=(1+2+3+4+...+n)+(1+2+3+4+...n)+(n+n+...n-times)\) \(S(2n)=S(n)+S(n)+n^2\) \(S(2n)=2S(n)+n^2\)
Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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15 Aug 2011, 03:59
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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]
Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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15 Aug 2011, 04:02
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!!
Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
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64% (02:09) correct 
kudos to u!!
