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Find the sum of n terms of the series (4-1/n)+(4-2/n)+(4-3/n)+.....
Asked by raginisood35 | 01 Jun, 2020, 08:24: PM Expert Answer
(4-1/n)+(4-2/n)+(4-3/n)+.....+(4-n/n)
= (4+4+...+upto n terms) - 1/n(1+2+3+...+n)
= 4n - 1/n[n(n+1)/2]
= 4n - (n+1)/2
= (8n - n - 1)/2
= (7n-1)/2
Answered by Renu Varma | 02 Jun, 2020, 11:36: AM

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