Question

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

Answer 1

We know that :
1 + 2 +3 + 4 +5 +6 + ...........+ n = $\frac{n(n+1)}{2}$
and
1 + 1+ 1 + 1 + 1 + .............+ n = n

Here,
$(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+\cdots$

= (4 + 4 + 4 + 4 + 4 + ......... upto n terms) + (-1/n - 2/n - 3/n - ..........upto n terms)

= 4 ( 1+1+1+1.......... upto n terms) - 1/n (1 + 2 + 3 +4 .........upto n terms)

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

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

= $\frac{8n-(n+1)}{2}$

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

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