# Find $\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)}$ [closed]

The question of course seems to be summation of a telescopic series, but despite many attempts am not able to reduce it ...

$$\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)}$$ is equal to ??

I've tried converting $2$'s to $(3-1)$ to reduce the expression, but that too didnt work out.

Edit:

I meant that the question seems to be one of 'telescopic method', but its not necessary to do it that way.

For confirmation or as a hint the answer is:

$$\frac{3}{2}-\frac{3^{101}}{101(102)}$$

Other might be useful information is that in this case,

$$\sum_{r=2}^{100}...=\sum_{r=1}^{100}...$$

## closed as unclear what you're asking by Did, Behrouz Maleki, Parcly Taxel, Leucippus, Daniel W. FarlowAug 26 '16 at 18:21

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• Using partial fraction decomposition $\frac{(2-2r)}{(r+1)(r+2)}=\frac{4}{r+1}-\frac{6}{r+2}$. Does this help ? – Claude Leibovici Aug 26 '16 at 8:56
• Note the $r$ in the denominator: $$\sum_{r=2}^{100}\frac{3^r(2-2r)}{r(r+1)(r+2)}=\sum_{r=2}^{100}\left(\frac{3^r}{r(r+1)}-\frac{3^{r+1}}{(r+1)(r+2)}\right)=\frac{3^2}{2\cdot3}-\frac{3^{101}}{101\cdot102}$$ – Did Aug 26 '16 at 10:59

With fraction decomposition we get $\frac{(2-2r)}{(r+1)(r+2)}=\frac{4}{r+1}-\frac{6}{r+2}$ (as mentioned by claude-leibovici), so \begin{align} \sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)} &= \sum_{r=2}^{100} 3^r \cdot (\frac{4}{r+1}-\frac{6}{r+2})\\ &= \sum_{r=2}^{100} \frac{3^r \cdot 4}{r+1}-\frac{3^r \cdot 6}{r+2} \\ &= \sum_{r=2}^{100} \frac{3^r \cdot 4}{r+1}-\frac{3^{r+1} \cdot 2}{r+2} \\ \end{align}

I am sure frome here you se the telescopic series.

• Nah, i got the same result but how to reduce that 4 thingy. – Display name Aug 26 '16 at 9:26
• "I am sure frome here you se the telescopic series." Hmmm, no I do not. (But the OP and at least two upvoters do, so we should be given an indication very soon). – Did Aug 26 '16 at 10:15

After the reduction to

$$S=\sum_{r=2}^{100}\left(\frac{4\cdot 3^{r}}{r+1}-\frac{2\cdot 3^{r+1}}{r+2}\right) = \sum_{r=2}^{100}\left(\frac{3^{r+1}}{r+1}-\frac{3^{r+2}}{r+2}\right)+\sum_{r=2}^{100}\left(\frac{3^r}{r+1}+\frac{3^{r+1}}{r+2}\right)$$ we have $$S = 12-\frac{2\cdot 3^{101}}{51}+2\cdot\sum_{r=3}^{99}\frac{3^r}{r+1}=\boxed{1-\frac{2\cdot 3^{100}}{17}+2\cdot\color{red}{\sum_{r=0}^{99}\frac{3^{r}}{r+1}}}$$ but the last term do not telescope (it is a rational number with a huge numerator and a huge denominator, by $p$-adic heights). Anyway, it is clearly related with $$\int_{0}^{3}\frac{x^{100}-1}{x-1}\,dx$$ that is not difficult to approximate, since the main contribute to the integral clearly comes from a neighbourhood of $x=3$. I guess there is a typo in the original exercise, the real telescopic sum is $$\sum_{r=2}^{100}\frac{3^r(\color{red}{2r+1})}{(r+1)(r+2)}=\sum_{r=2}^{100}\left(\frac{3^{r+1}}{r+2}-\frac{3^r}{r+1}\right) .$$

• Check out the edit – Display name Aug 26 '16 at 10:56
• @Displayname: check out the output of a CAS. The actual answer is given by a rational number whose denominator is $3951603758810864306182892650093458282400$, definitely not $101\cdot 102$. – Jack D'Aurizio Aug 26 '16 at 10:59
• It happens the typo is different, see comment on main. – Did Aug 26 '16 at 11:00
• @Did: I see, deeply flawed question here :/ – Jack D'Aurizio Aug 26 '16 at 11:01

Another approach is to use the geometric series starting from here: $$\sum_{r=2}^{100} x^{r-1}=\frac{1-x^{100}}{1-x}-1$$ and apply all needed transformations multipling/dividing by x and integrating and derivating until you get this: $$\sum_{r=2}^{100}\frac{x^r(2-2r)}{(r+1)(r+2)}=-\frac{2}{x^2} \int_0^x \left(\int_0^t w^2 \frac{\partial }{\partial w^1}\left(\frac{1-w^{100}}{1-w}-1\right) \, dw\right) \, dt$$ calculate the RHS since it's just 2 rational integrals and it can be done. then substitute $x \to 3$ and the job is done.

Remark: if you want to see all the steps just look at the RHS from the inside out and start off from the first formula in the post.

steps.

Der by x $\to$ Mul by $x^2$ $\to$ Int from $0$ to $x$ $\to$ Int from $0$ to $x$ $\to$ Mul by $-\frac{2}{x^2}$

• The double integral needs to be rewritten more rigorously (the same letter $x$ cannot be used for both integral bounds and as variables of integration). – Did Aug 26 '16 at 10:14