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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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Updated on: 17 Aug 2017, 05:53
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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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Originally posted by DeeptiM on 15 Aug 2011, 02:39.
Last edited by abhimahna on 17 Aug 2017, 05:53, edited 1 time in total.
Added OA.



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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, 03:22
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+ (n1)d = 1 + (n1) 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: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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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: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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15 Aug 2011, 04:46
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, \(S(n)=1+2+3+4,...+n\) \(S(2n)=1+2+3+4,...+n+(1+n)+(2+n)+(3+n)+(4+n),...+(n+n)\) \(S(2n)=(1+2+3+4+...+n)+(1+2+3+4+...n)+(n+n+...ntimes)\) \(S(2n)=S(n)+S(n)+n^2\) \(S(2n)=2S(n)+n^2\) Ans: "D"
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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, 04:59
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: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of
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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, \(S(n)=1+2+3+4,...+n\) \(S(2n)=1+2+3+4,...+n+(1+n)+(2+n)+(3+n)+(4+n),...+(n+n)\) \(S(2n)=(1+2+3+4+...+n)+(1+2+3+4+...n)+(n+n+...ntimes)\) \(S(2n)=S(n)+S(n)+n^2\) \(S(2n)=2S(n)+n^2\) Ans: "D" Thanks Fluke for saving my back on so many occasions kudos to u!!



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Re: 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
W.K.T \(S(n)=\frac{n(n+1)}{2}\)  first relation \(S(2n)=\frac{2n(2n+1)}{2}\) \(S(2n)=2n(\frac{n}{2}+\frac{n+1}{2})\) \(Substitute \frac{n+1}{2} = \frac{S(n)}{n} from1strelation\) \(S(2n)=2n(\frac{n}{2}+\frac{S(n)}{n})\) reduce \(S(2n) = n^2 + 2S(n)\) Ans : D
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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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12 Sep 2017, 12:34
S(n)=n(n+1)/2 S(n)=(n/2 )(n+1) S(2n)=(2n/2)(2n+1) S(2n)=n(2n+1) S(2n)=2n^2+n S(2n)=(n^2)+n +(n^2) S(2n)= 2S(n)+ n^2
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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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14 Sep 2017, 01:01
I calculated 2*S(n) in place of S(2*n)  silly mistake, and hence landed up in the answer option A.
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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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13 Aug 2018, 07:23
Hi, If S(n) = n(n+1)/ 2 or n^2+n= 2S(n) ....... (i) then S(2n)= 2n(2n+1)/2 = n(2n+1) = 2n^2 +n =n^2+n^2+n = n^2 + 2S(n) { substituting the value from eq i) Bunuel, Can we move this to PS forum.
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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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13 Aug 2018, 10:18
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 1. Number plugging approach: Let n = 3, we have S(n) = S(3) = 1+2+3 = 6 S(2n) = S(6) = 1+2+3+4+5+6 = 21 > D is the answer. 2. Mathematical approach:\(S(n) = 1+2+3+...+(n1)+n = \frac{(n+1)*n}{2}\) > \((n+1)*n = 2* S(n)\)\(S(2n) = 1+2+3+...+(2n1)+2n = \frac{(2n+1)*2n}{2}\) = \((2n+1)*n\) = \((n+1)*n\) + \(n*n\) = \(2*S(n)\) + \(n^2\) > Answer D. To moderatorsThis is a PS question. Please move it to PS subforum.
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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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13 Aug 2018, 11:46
Probus wrote: Hi, If S(n) = n(n+1)/ 2 or n^2+n= 2S(n) ....... (i) then S(2n)= 2n(2n+1)/2 = n(2n+1) = 2n^2 +n =n^2+n^2+n = n^2 + 2S(n) { substituting the value from eq i) Bunuel, Can we move this to PS forum. _____________ Done. Thank you.
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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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13 Aug 2018, 12:05
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 Given \(S(n) = 1 + 2 + 3 +.....+ n\) \(S(2n) = 1 + 2 + 3 +......+ n + (n+1) + (n+2) +.......+ (n+n)\) \(S(2n)  S(n) = (n+1) + (n+2) +.......+ (n+n) = n*n + (1 + 2 + 3 +....n)\) hence \(S(2n) = n^2 + 2S(n)\) Answer D. Thanks, GyM
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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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13 Jan 2019, 13:12
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 this question, i just assumed a series of 5 numbers 1,2,3,4,5 S(n) = 15 and n = 5 S(2n) = 55 ( 1,2,3,4,5,6,7,8,9,10) Now we need to find 55 when we substitute the values for n and S(n) in the answer options (A) 2*S(n) => 2* 15 =30 (B) n*S(n) => 5*15 =75 (C) 2n*S(n) => 10*15 =150 (D) 2S(n)+n^2 => 55 (E) S(n)+2n^2 Answer D
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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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