Multiple choice

The sum ${ S }_{ n }$ to $n$ terms of the series $\displaystyle \frac { 1 }{ 2 } +\frac { 3 }{ 4 } +\frac { 7 }{ 8 } +\frac { 15 }{ 16 } +...$ is equal to

  1. $\displaystyle { 2 }^{ n }-n-1$
  2. $\displaystyle 1-{ 2 }^{ -n }$
  3. $\displaystyle { 2 }^{ -n }+n-1$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The series is 1/2 + 3/4 + 7/8 + ... + (2^n - 1)/2^n. This can be written as (1 - 1/2) + (1 - 1/4) + ... + (1 - 1/2^n) = n - (1/2 + 1/4 + ... + 1/2^n). The sum of the geometric series is (1/2)(1 - (1/2)^n) / (1 - 1/2) = 1 - 2^-n. So the total sum is n - (1 - 2^-n) = n - 1 + 2^-n.

AI explanation

By observing the numerators (1, 3, 7, 15), we see the nth numerator is 2^n - 1, while the nth denominator is simply 2^n. We can rewrite the general nth term as (2^n - 1) / 2^n, which simplifies to 1 - 1/2^n, or 1 - 2^(-n). Summing this over n terms results in S_n = the sum of 1 from 1 to n minus the sum of (1/2)^n from 1 to n, which equals n - (1/2)(1 - (1/2)^n) / (1 - 1/2). Simplifying this expression yields S_n = n - 1 + 2^(-n), matching the formula 2^(-n) + n - 1.

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