Answer :

The given sum can be written as (1 + 1 + 1 + …) - (1/n, 2/n, 3/n, …)

Sum of first series up to n terms = 1 + 1 + 1 + … up to n terms


= n


Now, consider the second series:


Here, first term = a = 1/n


Common difference = d = (2/n) - (1/n) = (1/n)


Now, Sum of n terms of an arithmetic series is given by:


Sn = [2a + (n - 1)d]


Therefore sum of n terms of second arithmetic series is given by:


Sn = [2(1/n) + (n - 1)(1/n)]


= [(2/n) + 1 - (1/n)]


= [(1/n) + 1]


= = = (n + 1)/2


Now, sum of n terms of the complete series = Sum of n terms of first series - Sum of n terms of second series


= n - (n + 1)/2


= (2n - n - 1)/2


= 1/2 (n - 1)


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