Question Number 226515 by mr W last updated on 01/Dec/25
Commented by Ghisom_ last updated on 02/Dec/25
$${S}_{{k}} =\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\:\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}\:=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)} \\ $$$${S}_{\mathrm{100}} =\frac{\mathrm{5050}}{\mathrm{10101}} \\ $$
Commented by mr W last updated on 02/Dec/25
$${thanks}! \\ $$
Answered by peace2 last updated on 02/Dec/25
$${n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}=\left({n}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} =\left({n}^{\mathrm{2}} +\mathrm{1}−{n}\right)\left({n}^{\mathrm{2}} +\mathrm{1}+{n}\right) \\ $$$$\Sigma\frac{{n}}{\left({n}^{\mathrm{2}} +\mathrm{1}−{n}\right)\left({n}^{\mathrm{2}} +\mathrm{1}+{n}\right)}=\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}−{n}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}+{n}} \\ $$$$=\Sigma\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}+\left({n}−\mathrm{1}\right)}−\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}−{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\left({U}_{{n}−\mathrm{1}} −{U}_{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left({U}_{\mathrm{0}} −{U}_{\mathrm{100}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10101}}\right)=\frac{\mathrm{5050}}{\mathrm{10101}} \\ $$
Commented by mr W last updated on 02/Dec/25
$${thanks}! \\ $$
Answered by mr W last updated on 02/Dec/25
![Σ_(n=1) ^(100) (n/(n^4 +n^2 +1)) =Σ_(n=1) ^(100) (n/(n^4 +2n^2 +1−n^2 )) =Σ_(n=1) ^(100) (n/((n^2 +1)^2 −n^2 )) =Σ_(n=1) ^(100) (n/((n^2 −n+1)(n^2 +n+1))) =Σ_(n=1) ^(100) (1/2)((1/(n^2 −n+1))−(1/(n^2 +n+1))) =(1/2)Σ_(n=1) ^(100) [(1/((n−1)n+1))−(1/(n(n+1)+1))] =(1/2)[((1/(0×1+1))−(1/(1×2+1)))+((1/(1×2+1))−(1/(2×3+1)))+...+((1/(99×100+1))−(1/(100×101+1)))] =(1/2)((1/(0×1+1))−(1/(100×101+1))) =(1/2)(1−(1/(10101))) =((5050)/(10101))](https://www.tinkutara.com/question/Q226519.png)
$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{{n}}{{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}−{n}^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\left[\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)+\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{\mathrm{0}×\mathrm{1}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}+\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}+\mathrm{1}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{99}×\mathrm{100}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{100}×\mathrm{101}+\mathrm{1}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{0}×\mathrm{1}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{100}×\mathrm{101}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10101}}\right) \\ $$$$=\frac{\mathrm{5050}}{\mathrm{10101}} \\ $$