Question

Find the sum of n terms of the series (4-1/n)+(4-2/n)+(4-3/n)

Answer 1

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

Answer 2

Answer:

The sum of n terms series is $S_n=\frac{7n-1}{2}$

Step-by-step explanation:

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

To find : Find the sum of n terms of the series ?

Solution :

The Series $(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+...$ is an arithmetic series.

The first term is $a=4-\frac{1}{n}$

The common difference is $d=4-\frac{2}{n}-(4-\frac{1}{n})=-\frac{1}{n}$

The sum formula of A.P is given by,

$S_n=\frac{n}{2}[2a+(n-1)d]$

Substitute the value in the formula,

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

$S_n=\frac{n}{2}[2\times 4-\frac{2}{n}-\frac{n}{n}+\frac{1}{n}]$

$S_n=\frac{n}{2}[8-\frac{2}{n}-1+\frac{1}{n}]$

$S_n=\frac{n}{2}[7-\frac{1}{n}]$

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

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

Therefore, The sum of n terms series is $S_n=\frac{7n-1}{2}$