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