Method of differences summation

Question

I was trying some Cambridge past papers and it said to first separate into partial fractions and then find the sum of the sequence, however after splitting inot partial fractions I'm not getting the terms to cancel out like I normally do with these questions. Is there something I'm missing .been trying to manipulate the 3 sets of term but can't seem to get it . Thanks

$$\sum_{r=1}^n\frac4{r(r+1)(r+2)}$$

Answer 1

Take $$b_{r}=\frac{1}{r(r+1)}.$$

Then

$$b_{r}-b_{r+1}=\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}=\frac{2}{r(r+1)(r+2)}=\frac{1}{2}a_{r}.$$

So $$\sum_{r=1}^{n}a_r=\sum_{r=1}^{n}\frac{4}{r(r+1)(r+2)}= 2\sum_{r=1}^{n} \left(\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}\right)= 2\left(\frac{1}{2}-\frac{1}{(n+1)(n+2)}\right). $$

In fact, if you want to sum, via differences and telescopic summation, any finite series with a general term of the form

$$\frac{1}{(r+k)(r+k+1)...(r+m-1)(r+m)}$$

where $k,m$ arew integers with $k<m$, then take

$$b_{r}=\frac{1}{(r+k)(r+k+1)...(r+m-1)(r+m-1)}$$

Answer 2

Using calculus of finite differences with rising factorials,

Let $S_n = \sum_{r=1}^n\frac4{r(r+1)(r+2)} = \sum_{r=1}^n4r^{-\underline3}$

Then by integrating, we get $S_n = 4\frac{r^{\underline-2}}{-2} + C$

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

For $n = 1$, $S_n = 4/6 = C - 2/6 \implies C = 1$

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