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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]  15 Aug 2011, 01:39
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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]  15 Aug 2011, 02:22
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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]  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.

"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]  15 Aug 2011, 03:46
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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]  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]

so (n + n^2 ) equals 2*Sn and not just Sn.
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Re: Sequence is making me go bonkers!! [#permalink]  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"

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If S(n) is the sum of sequence 1, 2, 3, 4, ...n, in terms of [#permalink]  07 Sep 2014, 23:21
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   [#permalink] 07 Sep 2014, 23:21
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