Question Number 68611 by TawaTawa last updated on 14/Sep/19
Answered by mind is power last updated on 15/Sep/19
![(1/(r^2 −4))=(1/(4(r−2)))−(1/(4(r+2))) ⇒Σ_(r=3) ^(100) (1/(r^2 −4))=(1/4)Σ((1/(r−2))−(1/(r+2)))=(1/4)(1+(1/2)+(1/3)+(1/4)−(1/(99))−(1/(100))−(1/(101))−(1/(102))) x=25(1+(1/2)+(1/3)+(1/4)−(1/(99))−(1/(100))−(1/(101))−(1/(102)))=51.088.... [x]=51](https://www.tinkutara.com/question/Q68805.png)
$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}\left({r}−\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({r}+\mathrm{2}\right)} \\ $$$$\Rightarrow\sum_{{r}=\mathrm{3}} ^{\mathrm{100}} \frac{\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{{r}−\mathrm{2}}−\frac{\mathrm{1}}{{r}+\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{100}}−\frac{\mathrm{1}}{\mathrm{101}}−\frac{\mathrm{1}}{\mathrm{102}}\right) \\ $$$${x}=\mathrm{25}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{100}}−\frac{\mathrm{1}}{\mathrm{101}}−\frac{\mathrm{1}}{\mathrm{102}}\right)=\mathrm{51}.\mathrm{088}…. \\ $$$$\left[{x}\right]=\mathrm{51} \\ $$
Commented by TawaTawa last updated on 17/Sep/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$