Question Number 80362 by Power last updated on 02/Feb/20
Answered by key of knowledge last updated on 02/Feb/20
![(1/(1+2+...+i))−(1/(1+2+...+i+(i+1)))=(((1+...+i+(i+1))−(1+...+i))/((1+2+...+i)(1+2+...+i+(i+1))))= (((i+1))/((1+2+...+i)(1+2+...+i+(i+1)))) ⇒1−[((1/1)−(1/(1+2)))+((1/(1+2))−(1/(1+2+3)))+...+((1/(1+...+99))−(1/(1+...+100)))]= 1−[1−(1/(1+...+100))]=(1/(5050))](https://www.tinkutara.com/question/Q80381.png)
$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+…+\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)}=\frac{\left(\mathrm{1}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)\right)−\left(\mathrm{1}+…+\mathrm{i}\right)}{\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}\right)\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)\right)}= \\ $$$$\frac{\left(\mathrm{i}+\mathrm{1}\right)}{\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}\right)\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)\right)} \\ $$$$\Rightarrow\mathrm{1}−\left[\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{1}+…+\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{1}+…+\mathrm{100}}\right)\right]= \\ $$$$\mathrm{1}−\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+…+\mathrm{100}}\right]=\frac{\mathrm{1}}{\mathrm{5050}} \\ $$