For the sequence a1, a2, a3. ..an, an is defined by an=12n−1−12n for e

Question

For the sequence  a_1 ,  a_2 ,  a_3 . .. an , an is defined by  a_n=\frac{1}{2^{n−1}}−\frac{1}{2^n}  for each integer n≥1. What is the sum of the first 8 terms of the sequence?

A. 63/64
B. 127/128
C. 255/256
D. 511/512
E. 513/512

broall · Answer

a_1=\frac{1}{2^0}-\frac{1}{2^1} 
 a_2=\frac{1}{2^1}-\frac{1}{2^2} 
...
 a_8=\frac{1}{2^7}-\frac{1}{2^8}

a_1+a_2+...+a_8=\frac{1}{2^0}-\frac{1}{2^8}=\frac{255}{256} .

The answer is C.

Resad95 · Answer

I solved it in this way, but I think there can be less time consuming methods. (Anyway Bunuel would find a better way

) Rewrite equation as 2^{-(n-1)}-2^{-n} So we can take 2^{-n} out of parentheses and write as 2^{(-n)}( 2-1) , since 2^{(-n+1)} is 2^{(-n)}+2^1 We get ending up with 2^{(-n)} or \frac{1}{2^n} , put numbers from 1 to 8 \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}= \frac{255}{256} Actually you don't need to calculate numbers since there is only one answer choice with denominator of 256, Answer choice C Appreciate Kudos if this helped you, I put a great effort :wink:

Vinayak Shenoy · Answer

C
the series comes down to 1/2+ 1/4+ 1/8+....+1/256
taking common denominator and adding we get 255/256

warriorguy · Answer

My 2 cents:

The series is in geometric progression. The first term is 1/2 and Difference is 1/2.

Using formula; Sum =   /   = 255/256

C

Adigmat123 · Answer

Hi can you please explain how did you get 255/256 by taking common denominator and adding

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For the sequence a1, a2, a3. ..an, an is defined by an=12n−1−12n for e

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For the sequence a1, a2, a3. ..an, an is defined by an=12n−1−12n for e

Prep Toolkit

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My Rewards