Find the sum of (1 - 1/n) + (1 - 2/n) + (1 - 3/n) ... upto n terms

(n-1) / 2

Step-by-step explanation:

We have that the sum has the form of:

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

we compute the terms of a and d

a = 1 - 1 / n

a = (n - 1) / n

, that is, the first term, in the case d is the subtraction between the second and the first term, thus:

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

d = 1 - 2 / n - 1 + 1 / n

d = - 1 / n

now if replacing these values:

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

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

Sn = 1/2 * (n-1)

Therefore, the value of that sum is (n-1) / 2