Question

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

Answer 1

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