find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $

Question

$$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$

$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)} $$

I do not know how to get a telescoping series from here to cancel terms.

Answer 1

HINT:

Note that we have

$$\begin{align} \frac{2k-1}{k(k+1)(k+2)}&=\color{blue}{\frac{3}{k+1}}-\frac{5/2}{k+2}-\frac{1/2}{k}\\\\ &=\color{blue}{\frac12}\left(\color{blue}{\frac{1}{k+1}}-\frac1k\right)+\color{blue}{\frac52}\left(\color{blue}{\frac{1}{k+1}}-\frac{1}{k+2}\right) \end{align}$$

Answer 2

Let the fractions be $\frac{a}{k}$, $\frac{b}{k+1}$, and $\frac{c}{k+2}$.

$\frac{a}{k}+\frac{b}{k+1}+\frac{c}{k+2}=\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\frac{2k-1}{k(k+1)(k+2)}$

We want the following

$a+b+c=0$

$3a+2b+c=2$

$2a=-1$

Solve, $a=-\frac{1}{2}$, $b=3$, and $c=-\frac{5}{2}$.

The rest is standard.

Answer 3

When terms are in A.P in denominator, we use difference of last and first terms in product to distribute the terms of denominator over numerator, to change into telescoping series.

Answer 4

You are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $3$ or greater than $n$.

$$ \begin{aligned} & \sum_{k=1}^n\frac{-1}{2k} + \sum_{k=1}^n\frac{3}{k+1} - \sum_{k=1}^n\frac{5}{2}\frac{1}{k+2} \\ & = \left[-\frac{1}{2} - \frac{1}{4} + \frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{3}{2} + \frac{1}{2}\sum_{k=2}^n\frac{6}{k+1}\right] - \left[\frac{1}{2}\sum_{k=1}^n\frac{5}{k+2}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n+1}\frac{6}{k}\right] - \left[\frac{1}{2}\sum_{k=3}^{n+2}\frac{5}{k}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n}\frac{6}{k} + \frac{6}{2}\frac{1}{n+1}\right] - \left[\frac{1}{2}\sum_{k=3}^{n}\frac{5}{k} + \frac{5}{2}\frac{1}{n+1} + \frac{5}{2}\frac{1}{n+2}\right] \\ & = \frac{3}{4} + \frac{1}{2}\sum_{k=3}^{n}\left[\frac{-1 + 6 - 5}{k}\right] + \frac{6}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+2} \\ & = \frac{3}{4} + \frac{1}{2(n+1)} - \frac{5}{2(n+2)} \end{aligned} $$