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

Is there a way to answer this without having to sub in all 10 values for k? I have less than 120 seconds to answer questions like this on the GMAT exam...

Thank you!

Relevance

alternating series are bounded by their partial sums

In English that means if you have a sum like this: 1/2 - 1/4 + 1/8 -1/16 + 1/32 - 1/64 then the final answer is between any two "cut offs" so the final answer is between 1/2 and 1/2-1/4=1/4, the final answer is also between 1/2-1/4=1/4 and 1/2-1/4+1/8=3/8. The actual final answer to a+ar+ar^2+... (called a geomteric series) is a/(1-r) so in your problem the answer (when yous um to infinity) is (1/2)/(1-(-1/2))= (1/2)/(3/2) = 1/3. So the sum of the first 10 will be close to this final answer of 1/3

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• Sure.

The series is

1/2 - 1/4 + 1/8 - ...

Note that the series is alternating and the absolute value of each term is less than that of the preceeding term. So the final value is between the first term and the sum of the first two terms. That is to say, the sum is between 1/2 and (1/2 - 1/4 = 1/4). As each additional term is added, the sum stays within those bounds.

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• on the grounds that 9 is unusual 9* is 9 cases 3=27 on the grounds that 6* is even 6* is 6 cases a million/2=3 9* x 6* would desire to be 27X3=80 one Does the respond e book say 27* or 27? If it says 27*, 27* could be 27X3 as a results of fact it somewhat is unusual and could be comparable to an answer of 80 one. are additionally you specific the question did not ask for 9X6* which may well be 9x3=27? If neither of those are the case than i could agree including your answer. Is there any way you may submit a link to the positioning with the question on right here?

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• e

sub 1 and you get 1/4

sub 10 and you get a very samll negative fraction

thus every odd will give you a 1 * a smaller fraction then 1/4

every even will give you a negative so then the sum must be less then 1/4

there are always tricks to these problems to do them fast

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• This is - sum (1 to 10) (-1/2)^k

There is a formula for the geometric proggression

I think it's this one: sum (1 to n) (a)^k = [1-a^(n+1)]/(1-a)

Ana

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• the sum is

T = 1/2 - 1/4 + 1/8 - 1/16 ......... 1/(2^9) - 1/(2^10)

= 1/4 + 1/16 ...... 1/(2^10)

thus 1/4 < T < 1/2

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