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

"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?

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"

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.

Thanks Fluke for saving my back on so many occasions kudos to u!!

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

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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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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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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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:
\(S(n) = 1+2+3+...+(n-1)+n = \frac{(n+1)*n}{2}\)
--> \((n+1)*n = 2* S(n)\)
\(S(2n) = 1+2+3+...+(2n-1)+2n = \frac{(2n+1)*2n}{2}\) = \((2n+1)*n\) = \((n+1)*n\) + \(n*n\) = \(2*S(n)\) + \(n^2\)
--> Answer D.

To moderators
This is a PS question. Please move it to PS subforum.
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_____________
Done. Thank you.
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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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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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Very tough problem because it takes a lot of time to see the pattern. Are there other questions like this?
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