Math, asked by han18, 1 year ago

find the sum of n terms of the series (4-1/n)+(4-2/n)+(4-3/n)

Answers

Answered by saurabhsemalti
7

(4 -  \frac{1}{n} ) + (4 -  \frac{2}{n}) + ....(4 -  \frac{n}{n} ) \\  =  (4 + 4..n \: times) -  \frac{1}{n} (1 + 2 + ...n) \\  = 4n -  \frac{1}{n} ( \frac{n}{2} (2 + (n - 1)) \\  = 4n -  \frac{1}{2} (n + 1) \\ ....ans \\  \\  \\  \\  \\ or \\  \\  \\  =  \frac{8n - n - 1}{2}  \\  =  \frac{7n - 1}{2}
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