Find the sum of n terms of the following series:

Solution:

Let the given series be $X=\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots$

$=[4+4+4+\ldots]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots\right]$

$=4[1+1+1+\ldots]-\frac{1}{n}[1+2+3+\ldots]$

$=S_{1}-S_{2}$

$S_{1}=4[1+1+1+\ldots]$

$a=1, d=0$

$S_{1}=4 \times \frac{n}{2}[2 \times 1+(n-1) \times 0] \quad\left(S_{n}=\frac{n}{2}(2 a+(n-1) d)\right)$

$\Rightarrow S_{1}=4 n$

$S_{2}=\frac{1}{n}[1+2+3+\ldots]$

$a=1, d=2-1=1$

$S_{2}=\frac{1}{n} \times \frac{n}{2}[2 \times 1+(n-1) \times 1]$

$=\frac{1}{2}[2+n-1]$

$=\frac{1}{2}[1+n]$

Thus, $S=S_{1}-S_{2}=4 n-\frac{1}{2}[1+n]$

$S=\frac{8 n-1-n}{2}=\frac{7 n-1}{2}$

Hence, the sum of $n$ terms of the series is $\frac{7 n-1}{2}$.