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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

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

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  • $\begingroup$ What do you get after applying partial fractions? $\endgroup$ – Simply Beautiful Art Aug 4 '18 at 2:36
  • $\begingroup$ (2/r) - 4/(r+1) + 2/r+2 $\endgroup$ – user122343 Aug 4 '18 at 2:39
  • $\begingroup$ Can you figure out $\sum a_r-a_{r+1}$? What about if $a_r=\frac2r-\frac2{r+1}$? $\endgroup$ – Simply Beautiful Art Aug 4 '18 at 2:53
  • $\begingroup$ Yes I can do that since the terms cancel out easily $\endgroup$ – user122343 Aug 4 '18 at 2:56
  • $\begingroup$ Then you've figured out the answer? :-) $\endgroup$ – Simply Beautiful Art Aug 4 '18 at 3:02
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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

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

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  • $\begingroup$ thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ? $\endgroup$ – user122343 Aug 4 '18 at 3:25
  • $\begingroup$ @user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences. $\endgroup$ – Medo Aug 4 '18 at 3:33
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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 $S_n = 1 - \frac{2}{(n+1)(n+2)}$

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