Question Number 150597 by mathdanisur last updated on 13/Aug/21
Answered by Olaf_Thorendsen last updated on 13/Aug/21
$$\frac{\overline {{nnn}…{nn}}}{{n}+{n}+{n}…{n}}\:=\:\frac{{n}\underset{{p}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\mathrm{10}^{{p}} }{{kn}}\:=\:\frac{\frac{\mathrm{1}−\mathrm{10}^{{k}} }{\mathrm{1}−\mathrm{10}}}{{k}} \\ $$$$=\:\frac{\mathrm{10}^{{k}} −\mathrm{1}}{\mathrm{9}{k}} \\ $$$$\mathrm{S}\:=\:{x}+\underset{{n}=\mathrm{4}} {\overset{\mathrm{998}} {\sum}}\frac{\overline {{nnn}…{nn}}}{{n}+{n}+{n}+{n}…{n}} \\ $$$$\mathrm{S}\:=\:{x}+\underset{{n}=\mathrm{4}} {\overset{\mathrm{998}} {\sum}}\frac{\mathrm{10}^{{k}} −\mathrm{1}}{\mathrm{9}{k}}\:=\:{x}+\mathrm{995}\frac{\mathrm{10}^{{k}} −\mathrm{1}}{\mathrm{9}{k}} \\ $$$$\int\frac{{dx}}{\mathrm{S}}\:=\:\mathrm{ln}\mid{x}+\mathrm{995}\frac{\mathrm{10}^{{k}} −\mathrm{1}}{\mathrm{9}{k}}\mid+\mathrm{C} \\ $$
Commented by mathdanisur last updated on 14/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{cool} \\ $$