Question

Find the sum of the terms between the 102nd and the 190th terms in the series

$1/10 + 1/5 + 3/10 + 2/5 + 1/2 + ....$

Answer 1

$$a_1=\dfrac{1}{10}, a_2 =\dfrac{1}{5}=\dfrac{2}{10}, a_3=\dfrac{3}{10},$$

$$a_4=\dfrac{2}{5}=\dfrac{4}{10}, a_5 =\dfrac{1}{2}=\dfrac{5}{10}, ...$$

$$\dfrac{1}{10}+\dfrac{1}{5}+\dfrac{3}{10}+\dfrac{2}{5}+\dfrac{1}{2}+...=\dfrac{1}{10}\displaystyle\sum_{i=1}^\infin i$$

We know that

$$\displaystyle\sum_{i=1}^n i=\dfrac{n(n+1)}{2}$$

Then

$$\displaystyle\sum_{i=1}^{101} i=\dfrac{101(101+1)}{2}=5151$$

$$\displaystyle\sum_{i=1}^{190} i=\dfrac{190(190+1)}{2}=16235$$

$$\displaystyle\sum_{i=1}^{190} i-\displaystyle\sum_{i=1}^{101} i=16235-5151=11084$$

The sum of the terms between the 102nd and the 190th terms is

$$\dfrac{1}{10}(11084)=1108.4$$