Question

Find the sum of n terms of the series $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+..........$

Answer 1

Let the given series be S = $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+..........$
$$\begin{gathered} =\left[4+4+4+...\right]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...\right] \\ =4\left[1+1+1+...\right]-\frac{1}{n}\left[1+2+3+...\right] \\ =S_1-S_2 \end{gathered}$$
$$\begin{gathered} S_1=4\left[1+1+1+...\right] \\ a=1,d=0 \\ {S}_1=4\times \frac{n}{2}\left[2\times 1+\left(n-1\right)\times 0\right]\left(S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\right) \\ \Rightarrow S_1=4n \end{gathered}$$
$$\begin{gathered} S_2=\frac{1}{n}\left[1+2+3+...\right] \\ a=1,d=2-1=1 \\ {S}_2=\frac{1}{n}\times \frac{n}{2}\left[2\times 1+\left(n-1\right)\times 1\right] \\ =\frac{1}{2}\left[2+n-1\right] \\ =\frac{1}{2}\left[1+n\right] \end{gathered}$$
$$\begin{gathered} \mathrm{Thus},S=S_1-S_2=4n-\frac{1}{2}\left[1+n\right] \\ S=\frac{8n-1-n}{2}=\frac{7n-1}{2} \end{gathered}$$
Hence, the sum of n terms of the series is $\frac{7n-1}{2}$.