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Find the sum of \(1-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-...\)

 Jun 23, 2023
 #1
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The given series is an alternating geometric series, with first term 1 and common ratio -1/2. The sum of an alternating geometric series is equal to the first term divided by 1 + the absolute value of the common ratio, so the sum of the given series is

1 / (1 + (-1/2)) = 1 / (3/2) = 2/3.

Therefore, the answer is 2/3​.

 Jun 23, 2023
 #2
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The sequence can be re-written as:

 

1/4 + 1/32 + 1/ 256 + 1/2048 +..........

 

(1/32) / (1/4) ==1 / 8 - this is the common ratio.

 

Sum ==F / [1 - r], where F=First term,   r==Common ratio

 

Sum== 1/4 / [1  -  1/8] ==1/4 / (7/8)

 

Sum==(1/4) x (8/7) ==8 / 28 ==2 / 7

 Jun 23, 2023

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