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# What is the sum of n terms of the series 1/2 + 3/4 + 7/8 +

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What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + [#permalink]  06 Mar 2004, 15:51
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What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + 15/16 + ...equal to?

a. (2^n)-n-1
b. (2^-n)+n-1
c. (2^n)-1
d. (2^2n)-1
e.
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 30 Aug 2003
Posts: 330
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 1 [0], given: 0

reasoning pls
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
Followers: 1

Kudos [?]: 4 [0], given: 0

Quote:
What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + 15/16 + ...equal to?

a. (2^n)-n-1
b. (2^-n)+n-1
c. (2^n)-1
d. (2^2n)-1
e.

Series can be written as:
(2^n - 1) / (2^n) where n runs from 1 to n
Now sum of this series is:

=sum of(1-1/(2^n))
=sum of 1 - sum of (1/(2^n)),
where sum of 1 gives n and
Now, 1/(2^n) is geometric series with a=1/2 and r=1/2
gives sumation as,
a(1-r^n)/(1-r) = (1/2)(1-(1/2)^n)/(1-1/2) = 1 - 2^(-n)
n - (1 - 2 ^ (-n))
= 2^-n + n -1

That is B

Hope, i am clear enough

dharmin
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