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Sequence Q

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by gvosough » Mon Feb 14, 2011 2:45 pm
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) >2
b) 1<k<2
c) 1/2<k<1
d) 1/4<k<1/2
e)K<1/4

thanks,

OA: D
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by Night reader » Mon Feb 14, 2011 2:54 pm
simplify -1(k+1) * 2^-k
find T-?
for k=1 --> 1/2
for k=2 --> -1/4
for k=3 --> 1/8

Thus we can see that T will never be more than 1/2, as 1/2-1/4 and then all +ve values are less than 1/4
T can never be less than 1/4 as after (1/2 -1/4) we have only +ve totals (1/8-1/16...)
So pick D
gvosough wrote: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) >2
b) 1<k<2
c) 1/2<k<1
d) 1/4<k<1/2
e)K<1/4

thanks,

OA: D
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by vaflaly » Mon Feb 14, 2011 3:23 pm
Uk=((-1)^(K+1))*(1/2^K)
Uk= (-1) (-1)^K * (1/2^K)
Uk= - (-1/2)^k

u1=-1/2 u2=1/4 u1+u2= 1/4

u3=-1/8 u4=1/16 u3+u4=1/16

u5=-1/32 u6=1/64 u5+6=1/64

so, u7+u8=1/256 ; u9+u10=1/1024

Sum(1->10) = 1/4+1/16+1/64+1/1024
Sum(1->10) = 436/1024
Sum(1->10) = 0,33

Answer is D
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by garuhape » Mon Feb 14, 2011 5:03 pm
The only thing they ask you to do is to give the range of the accumulated ks.

For k=1,3,5,7,9 you get positive values which are decreasing/converge to zero.
For k=2,4,6,8,10 you get negative values which are increasing/converge to zero.

Therefore k=1 is the greatest value and k=2 is the smallest value [0.25<k<0.5]

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