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For every integer K (Exam on Friday!)

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For every integer K (Exam on Friday!)

by ssuarezo » Wed Jun 16, 2010 5:18 pm
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is (-1)^(k+1)*(1/(2^k)). If T is the sum of the first 10 terms, then T is:

a- greater than 2
b- betwn 1 and 2
c- betwn 1/2 and 1
d- betwn 1/4 and 1/2
e- less than 1/4

oa is d

I attach file in case the seq is not clear. (1/(2^k)) multiplies (-1), not its exponent

Thanks
Silvia
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by amising6 » Wed Jun 16, 2010 5:28 pm
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is (-1)^(k+1)*(1/(2^k)). If T is the sum of the first 10 terms, then T is:

a- greater than 2
b- betwn 1 and 2
c- betwn 1/2 and 1
d- betwn 1/4 and 1/2
e- less than 1/4

if we solve for only first few terms
t(1)=+1/2 (putting k=1 in kth ter sequence given)
t(2)=-1/4
t(3)=1/8

t(4)=-1/16
so you can say all odd will be positive and all even term will be negative
if you take only t(1) and t(2)=+1/2 - 1/4=1/4
now going fore ward t(3)and t(4)=+1/8 - 1/16=1/16
t(5) and t(6)=1/32
t(7) and t(8)=1/64

so on and so far
so you can say number will be more than 1/4 as only t(1) and t(2) gives you 1/4
remaining terms will contribute something but og not great contribution so you can say it will be betwn 1/4 and 1/2
Ideation without execution is delusion
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by ssuarezo » Wed Jun 16, 2010 5:56 pm
amising6 wrote:if we solve for only first few terms
t(1)=+1/2 (putting k=1 in kth ter sequence given)
t(2)=-1/4
t(3)=1/8

t(4)=-1/16
so you can say all odd will be positive and all even term will be negative
if you take only t(1) and t(2)=+1/2 - 1/4=1/4
now going fore ward t(3)and t(4)=+1/8 - 1/16=1/16
t(5) and t(6)=1/32
t(7) and t(8)=1/64

so on and so far
so you can say number will be more than 1/4 as only t(1) and t(2) gives you 1/4
remaining terms will contribute something but og not great contribution so you can say it will be betwn 1/4 and 1/2
Thank you Amising6
Actually, I got the numbers, but not the conclusion (I started summing it all up!).
Silvia.
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