Question 41

The sum $$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + ....... + \frac{1023}{1024}$$ lies between

The series can be broken as: $$\ \frac{\ 1}{2}+\ \ \frac{\ 3}{4}+\ \frac{\ 7}{8}+.......\ \ \frac{\ 1023}{1024}$$ or $$\left(1-\ \frac{\ 1}{2}\right)+\left(\ 1-\ \frac{\ 1}{4}\right)+......$$

This becomes: 1+1+1+1..... n times - { $$\ \frac{\ 1}{2}+\ \frac{\ 1}{2^2}+.......\ \ \frac{\ 1}{2^n}$$ }

Here, n = 10

Hence, the sum becomes 10 - (Sum of GP upto 10 terms with a common ratio of 0.5)

This gives the answer as: 10 - $$\ \frac{\ 1}{2}\times\ \ \frac{\ \ 1-\frac{\ 1}{2^{10}}}{1-\ \frac{\ 1}{2}}$$ or approximately 9 (slightly greater than 9)

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