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

Question

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

PiyushK · Accepted Answer

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

viks4gmat · Answer

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  / 2 = n + 2n^2 = n + n^2 + n^2

we know from (1)

S(2n) = 2Sn + n^2 hence answer D

meshell · Answer

"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?

fluke · Answer

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"

viks4gmat · Answer

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 =   / 2 and is therefore indeed sn =   / 2
but to avoid confusion, i have pulled the 2 to the other side making it 2* Sn =

so (n + n^2 ) equals 2*Sn and not just Sn.

viks4gmat · Answer

Thanks Fluke for saving my back on so many occasions :-)

kudos to u!!

AbhimanyuDhar · Answer

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

AD2GMAT · Answer

I calculated 2*S(n) in place of S(2*n) - silly mistake, and hence landed up in the answer option A.

Probus · Answer

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.

Izzyjolly · Answer

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.

Bunuel · Answer

_____________
Done. Thank you.

GyMrAT · Answer

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

KanishkM · Answer

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

CEdward · Answer

Very tough problem because it takes a lot of time to see the pattern. Are there other questions like this?

Tiffsootmoot8 · Answer

Keep it simple..
Substitute any value for n and calculate S(n) and S(2n)
For example, for n=4, S(4) = 10 and S(8) = 36.
Now try substituting n=4 for each of the options and see which one matches. 
Only D works!

bumpbot · Answer

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

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