Question

The sum of the series $(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+\cdots$ upto $n$ terms is

• A. $\frac{5n+1}{2}$
• B. $\frac{7n-1}{2}$
• C. $\frac{7n+1}{2}$
• D. $\frac{5n-1}{2}$

Answer 1

The correct option is B $\frac{7n-1}{2}$

Given:

$S=(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+\cdots =(4+4+4+...n\text{times})-\frac{1}{n}(1+2+3+....n\text{times})$

We know, $\sum _{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$ (sum of first n natural numbers)

$\Rightarrow S=4n-\frac{1}{n}\left(\frac{n(n+1)}{2}\right)$

$\Rightarrow S=4n-\frac{n+1}{2}$

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

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

Hence, option B.