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apollo168
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For every k from 1 to 10 inclusive, the kth term of a 03 Sep 2006, 02:47
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For every 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 1st ten terms then T=?

Looking for a faster way to solve this problem. Wasted a lot of my time on this one number Thanks



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netskater
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For every k from 1 to 10 inclusive, the kth term of a 03 Sep 2006, 11:04
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apollo168
For every 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 1st ten terms then T=?

Looking for a faster way to solve this problem. Wasted a lot of my time on this one number Thanks
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The kth term is given by:

a(k) = (-1)^(k+1) (1/2)^k

By taking (-1)^k inside and separating -1 we can rewrite this as:

a(k) = - (-1/2)^k

Thus the series of terms is like: -1,1/4, -1/8, 1/16 and so on
First term, a = -1, common ratio, r = -1/2
Using the formula:
T = a * (1-r^10)/(1-r)
= -1 (1- (1/2)^10)/(1-(-1/2)
(1/2)^10 can be ignored as it is too small

T = -1/(3/2) = -2/3
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apollo168
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For every k from 1 to 10 inclusive, the kth term of a 03 Sep 2006, 11:12
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netskater
apollo168
For every 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 1st ten terms then T=?

Looking for a faster way to solve this problem. Wasted a lot of my time on this one number Thanks

The kth term is given by:

a(k) = (-1)^(k+1) (1/2)^k

By taking (-1)^k inside and separating -1 we can rewrite this as:

a(k) = - (-1/2)^k

Thus the series of terms is like: -1,1/4, -1/8, 1/16 and so on
First term, a = -1, common ratio, r = -1/2
Using the formula:
T = a * (1-r^10)/(1-r)
= -1 (1- (1/2)^10)/(1-(-1/2)
(1/2)^10 can be ignored as it is too small

T = -1/(3/2) = -2/3
Show more


Thanks never bothered to memorize the sum of the geometric sequence. But shouldnt the first term be 1/2?



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!