Question Number 183690 by paul2222 last updated on 28/Dec/22
Answered by mr W last updated on 29/Dec/22
![T_n =(n/a_n ) a_n =1^6 +2^6 +3^6 +...+n^6 =((n(n+1)(2n+1)(3n^4 +6n^3 −3n+1))/(42)) T_n =((42)/((n+1)(2n+1)(3n^4 +6n^3 −3n+1))) S=Σ_(n=1) ^∞ ((42)/((n+1)(2n+1)(3n^4 +6n^3 −3n+1))) =42Σ_(n=1) ^∞ [((45n^3 +21n^2 −57n+30)/(31(3n^4 +6n^3 −3n+1)))+((32)/(31(2n+1)))−(1/(n+1))] .....](https://www.tinkutara.com/question/Q183724.png)
$${T}_{{n}} =\frac{{n}}{{a}_{{n}} } \\ $$$${a}_{{n}} =\mathrm{1}^{\mathrm{6}} +\mathrm{2}^{\mathrm{6}} +\mathrm{3}^{\mathrm{6}} +…+{n}^{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}^{\mathrm{4}} +\mathrm{6}{n}^{\mathrm{3}} −\mathrm{3}{n}+\mathrm{1}\right)}{\mathrm{42}} \\ $$$${T}_{{n}} =\frac{\mathrm{42}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}^{\mathrm{4}} +\mathrm{6}{n}^{\mathrm{3}} −\mathrm{3}{n}+\mathrm{1}\right)} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{42}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}^{\mathrm{4}} +\mathrm{6}{n}^{\mathrm{3}} −\mathrm{3}{n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:=\mathrm{42}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{45}{n}^{\mathrm{3}} +\mathrm{21}{n}^{\mathrm{2}} −\mathrm{57}{n}+\mathrm{30}}{\mathrm{31}\left(\mathrm{3}{n}^{\mathrm{4}} +\mathrm{6}{n}^{\mathrm{3}} −\mathrm{3}{n}+\mathrm{1}\right)}+\frac{\mathrm{32}}{\mathrm{31}\left(\mathrm{2}{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right] \\ $$$$….. \\ $$
Commented by paul2222 last updated on 30/Dec/22
$$\boldsymbol{{W}}{ow} \\ $$