Question Number 207845 by hardmath last updated on 28/May/24
$$\mathrm{Find}: \\ $$$$\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\:+...+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{40}} \\ $$
Answered by Frix last updated on 28/May/24
$$\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{j}} {\sum}}{k}\right)=\mathrm{2}−\frac{\mathrm{2}}{{n}+\mathrm{1}} \\ $$$${n}=\mathrm{40}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{80}}{\mathrm{41}} \\ $$
Commented by hardmath last updated on 28/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Answered by MM42 last updated on 28/May/24
![=(2/(1×2))+(2/(2×3))+(2/(3×4))+...+(2/(39×40))+(2/(40×41)) =2[((1/1)−(1/2))+((1/2)−(1/3))+...+((1/(39))−(1/(40)))+((1/(40))−(1/(41)))] =2(1−(1/(41)))=((80)/(41)) ✓](Q207851.png)
$$=\frac{\mathrm{2}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}×\mathrm{4}}+...+\frac{\mathrm{2}}{\mathrm{39}×\mathrm{40}}+\frac{\mathrm{2}}{\mathrm{40}×\mathrm{41}} \\ $$$$=\mathrm{2}\left[\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+...+\left(\frac{\mathrm{1}}{\mathrm{39}}−\frac{\mathrm{1}}{\mathrm{40}}\right)+\left(\frac{\mathrm{1}}{\mathrm{40}}−\frac{\mathrm{1}}{\mathrm{41}}\right)\right] \\ $$$$=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{41}}\right)=\frac{\mathrm{80}}{\mathrm{41}}\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 28/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$