Find the sum of 𝑛 terms of the series

$(4-\frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n})+.....$

We need to find the sum of 𝑛 terms of the series

$(4-\frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n})+.....$

It can be observed that the series forms an AP with common difference

$(4 - \frac{2}{n}) - (4 - \frac{1}{n}) = -\frac{1}{n}$

It is known that the sum of an AP is given by

𝑆_n  = 𝑛/2[2𝑎 + (𝑛 − 1)𝑑], where 𝑆_n is the sum of terms, 𝑛 is the number of terms, 𝑎 is the first term and 𝑑 is the common difference of AP. On substituting the values, we get
$$\begin{gathered} S_{n = }\frac{n}{2}\left[2(4 -\frac{1}{n})+(n-1)(-\frac{1}{n})\right] \\ {S}_n = \frac{n}{2}\left[8 - \frac{2}{n}- 1 + \frac{1}{n}\right] \\ {S}_n = \frac{n}{2}\left[7-\frac{1}{n}\right] \end{gathered}$$

Hence, the sum of the series is $\frac{n}{2}\left[7 -\frac{1}{n}\right]$