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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of

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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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New post Updated on: 17 Aug 2017, 05:53
2
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A
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D
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Question Stats:

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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

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 [#permalink]

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New post 15 Aug 2011, 03:22
1
2
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: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]

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New post 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 [#permalink]

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New post 15 Aug 2011, 04:46
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,
\(S(n)=1+2+3+4,...+n\)
\(S(2n)=1+2+3+4,...+n+(1+n)+(2+n)+(3+n)+(4+n),...+(n+n)\)
\(S(2n)=(1+2+3+4+...+n)+(1+2+3+4+...n)+(n+n+...n-times)\)
\(S(2n)=S(n)+S(n)+n^2\)
\(S(2n)=2S(n)+n^2\)

Ans: "D"
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Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]

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New post 15 Aug 2011, 04:59
1
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 [#permalink]

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New post 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"



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Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]

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New post 08 Sep 2014, 00:21
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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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Re: If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]

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New post 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 [#permalink]

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New post 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   [#permalink] 14 Sep 2017, 01:01
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