Find the sum of *n* terms of the series `(4 - 1/n) + (4 - 2/n) + (4 - 3/n)+ ......`

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#### Solution

Let the given series be S = `(4 - 1/n) + (4 - 2/n) + (4 - 3/n)+ ......`

`= [4 + 4 + 4 + ...]- [1/n + 2/n + 3/n + ....]`

`= 4[1 + 1 + 1 + ...] - 1/n[1 + 2 + 3 + ...]`

`= S_1 - S_2`

`S_1 = 4[1 + 1 + 1 + ...]`

a = 1, d = 0

`S_1 = 4 xx n/2 [2 xx 1 + (n -1)xx 0]` `(S_n = n/2(2a + (n-1)d))`

`=> S_1 = 4n`

`S_2 = 1/n [1 + 2 + 3 + ....]`

a = 1, d = 2 -1 = 1

`S_2 = 1/n xx n/2[2 xx 1 + (n -1) xx 1]`

`= 1/2[2 + n - 1]`

=`1/2[1 + n]`

Thus `S = S_1 - S_2 = 4n - 1/2 [1 + n]`

`S = (8n - 1 - n)/2 = (7n - 1)/2`

Hence the sum of the n terms of the series is `(7n - 1)/2`

Concept: nth Term of an AP

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