Question Number 218651 by hardmath last updated on 13/Apr/25
Answered by vnm last updated on 14/Apr/25
$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{4}}\right)+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{8}}\right)+ \\ $$$$…+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}{n}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{4}{n}}\right)= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{{k}}−\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{k}}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{{k}}−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}= \\ $$$$\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$
Commented by hardmath last updated on 17/Apr/25
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{professor} \\ $$