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Mathematics Question on Sequence and series

If Hn=1+12+.....+1n,H_n = 1+\frac{1}{2} + ..... + \frac{1}{n}, then the value of Sn=1+32+53+.....+2n−1nisS_n = 1+\frac{3}{2} + \frac{5}{3} + .....+\frac{2n-1}{n} is

A

Hn+2nH_n + 2n

B

n−1+Hnn-1 + H_n

C

Hn−2nH_n - 2n

D

2n−Hn2n - Hn

Answer

2n−Hn2n - Hn

Explanation

Solution

Given, Hn=1+12+...+1nH_n = 1 + \frac{1}{2} + ...+\frac{1}{n}
Also, Sn=1+32+53+...+2n−1nS_n = 1 + \frac{3}{2} + \frac{5}{3} + ...+ \frac{2n-1}{n}
=1+(4−1)2+(6−1)3+...+2n−1n= 1 + \frac{(4-1)}{2} + \frac{(6 - 1)}{3} + ...+\frac{2n-1}{n}
=1+2−12+2−13+...+2−1n= 1 + 2 - \frac{1}{2} + 2 - \frac{1}{3} + ...+ 2 - \frac{1}{n}
=1+2(n−1)−(12+13+...+1n)= 1 + 2(n-1) - (\frac{1}{2} + \frac{1}{3} + ...+ \frac{1}{n})
=1+2(n−1)−(Hn−1)= 1 + 2(n-1) - (H_n - 1)
=2n−Hn= 2n - H_n