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Mathematics Question on Sum of First n Terms of an AP

Sum of the first n terms of the series 12+34+78+1516+...\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + ... is equal to

A

2n−n−12^n - n - 1

B

1−2n1 - 2^n

C

n+2n−1 n + 2^n - 1

D

2n+12^n + 1

Answer

n+2n−1 n + 2^n - 1

Explanation

Solution

Sum of the n terms of the series 12+34+78+1516+...\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + ... upto n terms can be written as
(1−12)+(1−14)+(1−18)+(1−116)+... upto n terms\bigg(1-\frac{1}{2}\bigg) + \bigg(1-\frac{1}{4}\bigg) + \bigg(1-\frac{1}{8}\bigg) + \bigg(1-\frac{1}{16}\bigg) + ... \, upto\, n\, terms
=n−(12+14+18+...+n terms)= n -\bigg(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + n\, terms \bigg)
n−12(1−12n)1−12=n+2−n−1n -\frac{\frac{1}{2}\bigg(1-\frac{1}{2^n} \bigg) }{1-\frac{1}{2}} = n + 2^{-n}-1