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

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2n−n−1

1−2−n

2−n+n−1

2n−1

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detailed solution

Correct option is C

We have Sn=1−12+1−14+1−123+…+1−12n=n−12+122+…+12n=n-121−12n1−12=n−1+2−n

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The sum Sn to n terms of the series 12+34+78+1516+…is equal to

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