Question

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

Answer 1

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

The sum upto $n$ terms is

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

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

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

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

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

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