Series JEE 2

If we observe the denominators, $2 , 5 , 9 , 14 , 20 , \cdots\cdots$ We can see that the difference between two consecutive is in an A.P. with common difference 1. Hence we can find the $n^{th}$ term. $n^{th} term = T_{n} = \dfrac{(n+1)(n+4)}{2}$ So, $n^{th}$ term of given sequence is $\dfrac {1}{T_{n}}$ . To get first term, put $n =0$ ,

To get second term put $n=1$ , and so on.

Now we have to find sum of first $100$ terms, i.e. $\displaystyle\sum_{n=0}^{99} \dfrac{1}{T_n}$ $= \displaystyle\sum_{n=0}^{99} \dfrac{2}{(n+1)(n+4)}$ $= \dfrac{2}{3} \displaystyle\sum_{n=0}^{99} ( \dfrac{1}{n+1} - \dfrac{1}{n+4} )$

All of the terms cancel out except the first three terms, and three terms at the end which are essentially 0. Thus, we have $\dfrac{2}{3} (1 + \dfrac{1}{2} +\dfrac{1}{3})$ $= \boxed{1.2222222\ldots}$

$2(\frac{1}{1.4}+\frac{1}{2.5}+\frac{1}{3.6}+\frac{1}{4.7}+\frac{1}{5.8}+\ldots)$

= $\displaystyle \sum_{r=1}^{n}\frac{2}{r(r+3)}$

= $\displaystyle \frac{2}{3}(\sum_{r=1}^{n}\frac{1}{r}-\frac{1}{r+3})$

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

= $\frac{n(11n^{2}+48n+49)}{9(n+1)(n+2)(n+3)}$

For large $n$ , we just look at the $n^3$ term, so it is $\frac{11}{9} = 1.2222\ldots.$

$\begin{aligned} & = \frac {1}{1 \times 2} + \frac {1}{1 \times 5} + \frac {1}{3 \times 3} + \frac {1}{2 \times 7} + \ldots \\ & = \displaystyle \sum^{\infty}_{n=1}\frac{1}{\frac{n(n+1)}{2}+n} = \displaystyle \sum^{\infty}_{n=1} \frac{2}{n^2+3n} \\ & = 2 \displaystyle \sum^{\infty}_{n=1} \frac{1}{n(n+3)} = \frac{2}{3} \displaystyle \sum^{\infty}_{n=1} \left(\frac{1}{n}-\frac{1}{n+3} \right) \\ & = \frac{2}{3} \left( \displaystyle \sum^{\infty}_{n=1}\frac{1}{n}- \displaystyle \sum^{\infty}_{n=4}\frac{1}{n}\right) \\ & = \frac{2}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right) \\ & = \frac{2}{3}\cdot \frac{11}{6} = \frac{11}{9} = \boxed{1.22} \end{aligned}$

Call the sum as $S_n$ . Then $\begin{aligned} S_n & =\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+\frac{1}{14}\cdots\\& = \frac{1}{3-1}+\frac{1}{6-1}+\frac{1}{10-1}+\frac{1}{15-1}+\cdots\\& = \displaystyle\sum_{n=2}^{\infty}\frac{2}{n(n+1)-2} \\& = \displaystyle\sum_{n=2}^{\infty}\frac{2}{(n-1)(n+2)} \\& = \frac{2}{3}\displaystyle\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+2}\right) \\&= \frac{2}{3}\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+3}\right)\\& = \frac{2}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right) \\& = \frac{11}{9} \\& =\boxed{1.2222} \end{aligned}$