Question

Find the sum
4-$4-\frac{1}{n} + 4-\frac{2}{n} + 4-\frac{3}{n} upto n terms$

Answer 1

Here wo goooooooo....

$4 +( 4 - \frac{1}{n} ) + (4 - \frac{2}{n} ) + (4 - \frac{3}{n}) .... \\ \\ = 4n - \frac{1}{n} - \frac{2}{n} - \frac{3}{n} .... \\ \\ = 4n - ( \frac{1}{n} + \frac{2}{n} ..) \\ = 4n - \frac{1}{n} (1 + 2 + 3....) \\ \\ = 4n - (\frac{1}{n} \times \frac{n(n + 1)}{2} )\\ = 4n - \frac{(n + 1)}{2} \\ = \frac{8n - n - 1}{2} \\ = \frac{7n - 1}{2}$

Therefore the sum of n terms will be:
$\frac{7n - 1}{2}$

Answer 2

Answer:

Step-by-step explanation: