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# Find the Sum of N Terms of the Series (4 - 1/N) + (4 - 2/N) + (4 - 3/N)+ ...... - Mathematics

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

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