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Integer - gmatprep

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Integer - gmatprep

by arocks » Fri Oct 12, 2007 8:46 am
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by
(-1)^k+1*(1/2^K). If T is the sum of the first 10 terms in the sequence, then T is

A) greater than 2
B) between 1 and 2
C) between 1/2 and 1
D) between 1/4 and 1/2
E) less than 1/4

Please explain. Thanks.

OA-D
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by Suyog » Fri Oct 12, 2007 9:34 am
here the first term is (-1) ^ k+1

for k = 1;
when you start calculating it the first number will be -1^2 = 1
and the (1/2)^k i.e. 1/2
so first term will be 1*1/2= 1/2

when k = 2;
(-1) ^ k+1 will be -1
and (1/2) ^ k = 1/4
so the second term will be -1 * 1/4 = -1/4
the sequence will continue... the next number will be positive and the number after that will be negative.

1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + 1/128 .... - 1/1024

lokking the this my guess was ans between 1/2 and 1/4; as the numbers ahead are getting more and more negligible.

you can select E. less than 1/4; but intelligent guess will be between 1/4 and 1/2;

if you go on and calculate the sequence; it gives 341/1024 = 0.33 i.e. approx 1/3

Ans D
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by arocks » Fri Oct 12, 2007 10:41 am
makes sense...thanks.
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by ldoolitt » Sun Oct 14, 2007 7:05 am
A slightly simpler way of looking at it.

The terms will alternate between positive and negative. Pair up each positive and negative term, and look at the pattern.

(1/2 – 1/4) + (1/8 – 1/16) + …
=1/4 + 1/16

1/4 is large, and every term being added after that is decreasingly smaller. Thus the sum will be greater than 1/4, but not by much. Which leaves (D).
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by jangojess » Sun Oct 14, 2007 10:33 pm
given seq = -1^k + (1/2)^k...now the term -1^k will be -1 for K with odd value and +1 for K with even value...in the 1st 10 numbers we'll have alternate -1 and +1 which will sum to ZERO...guys this is the main catch..we're then left with sum of (1/2)^k...now as the seq goes to infinity the value nears (1/2)/(1-1/2) = 1..so for 1st ten terms the sum shld be b/w 1/2 and 1...which is C.
Trying hard!!!
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