Question Number 9600 by Sopheak last updated on 20/Dec/16
Commented by prakash jain last updated on 20/Dec/16
$${k}^{{th}} \:\mathrm{Term}=\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\mathrm{18}×\left(\mathrm{10}^{\mathrm{2}{i}} \right)=\frac{\mathrm{18}\left(\mathrm{10}^{\mathrm{2}{k}} −\mathrm{1}\right)}{\mathrm{99}} \\ $$$${S}_{{n}} =\frac{\mathrm{2}}{\mathrm{11}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{10}^{\mathrm{2}{k}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{11}}\left(\frac{\mathrm{10}^{\mathrm{2}} \left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\mathrm{99}}−{n}\right) \\ $$