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Old 10-05-2008, 11:21 AM   #1 (permalink)
mehrak
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If S(n) is the sum of sequence

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

SPOILER:
OA: D
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Old 10-05-2008, 11:34 AM   #2 (permalink)
e.cartman
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S(n)=n(n+1)/2
S(2n) = 2n(2n+1)/2 = n(2n+1) = n^2+n(n+1) = n^2+2S(n).
So D.
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Old 10-07-2008, 02:03 PM   #3 (permalink)
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Quote:
Originally Posted by e.cartman View Post
S(n)=n(n+1)/2
S(2n) = 2n(2n+1)/2 = n(2n+1) = n^2+n(n+1) = n^2+2S(n).
So D.
excellent explanation e.cartman!
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Old 10-07-2008, 09:42 PM   #4 (permalink)
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For those who dont remember the formula:

Put n=3 and check the choices. Only D fits in. Hit and Trial
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Old 11-18-2008, 07:51 AM   #5 (permalink)
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Quote:
Originally Posted by e.cartman View Post
S(n)=n(n+1)/2
S(2n) = 2n(2n+1)/2 = n(2n+1) = n^2+n(n+1) = n^2+2S(n).
So D.
cartman, could you please explain it further how you arrived at

n(2n+1) = n^2+n(n+1) = n^2+2S(n)
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Old 11-18-2008, 08:18 AM   #6 (permalink)
swethapv
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The sum of an arithmetic series is given by the formula:
n(a1 + an)/2
Where, n is the number of numbers in the series
a1 is the 1st number in the series
an is the last number of the series
Now we can find an by the formula:
an = a1 + (n-1)d
where d is the difference between two consecutive numbers in the series
therefore, our sum formula becomes (by replacing an):
S(n) = n[2a1 + (n-1)d]/2
Now in the given series,
a1 = 1
d = 1
hence the equation becomes:
S(n) = n[2 + n-1]/2 = n(n+1)/2 =(n^2 + n)/2
therefore S(2n) = 2n^2 + n
so check each of the choices to see which one of them yields the above equation. Only choice D satisfies the above equation.
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Old 11-18-2008, 09:19 AM   #7 (permalink)
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Smile

Quote:
Originally Posted by Gmater-1 View Post
cartman, could you please explain it further how you arrived at

n(2n+1) = n^2+n(n+1) = n^2+2S(n)
sum of n terms = n(n+1)/2

we need to find S(2n). put 2n in place of n

S(2n) = 2n(2n+1)
hence D is the answer
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Old Yesterday, 05:47 PM   #8 (permalink)
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Quote:
Originally Posted by vinhan View Post
sum of n terms = n(n+1)/2

we need to find S(2n). put 2n in place of n

S(2n) = 2n(2n+1)
hence D is the answer
Still not clear "2n(2n+1)" is not in any of the answer choices
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Old Yesterday, 06:32 PM   #9 (permalink)
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S(n) =n(n+1)/2
So, S(2n) =2n(2n+1)/2 =2n^2+n =n^2+n^2+n =n^2+n(n+1)
=n^2+2n(n+1)/2 [Multiplying and diving by 2]
=n^2+2S(n) [As S(n)=n(n+1)/2]
Answer: n^2+2S(n)
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Old Yesterday, 07:48 PM   #10 (permalink)
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Quote:
Originally Posted by Gmater-1 View Post
Still not clear "2n(2n+1)" is not in any of the answer choices
That would be too easy...they like to make the process simple enough so that you can arrive to the solution, but then have to pass through five pains in order to understand which of the answers, transformed, is equal to it.
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