Question

Fie a=$\frac{1}{1*2}$+$\frac{1}{2*3}$+$\frac{1}{3*4}$+..............+$\frac{1}{99*100}$. Demonstrati ca $(a+0,01)^{2018}$ ∈ ℕ.[tex][/tex]

Answer 1

Știm că $\frac{1}{n(n+1)} =\frac{1}{n}-\frac{1}{n+1}$

Atunci $a=\frac{1}{1*2} +\frac{1}{2*3}+...+\frac{1}{99*100}= 1-\frac{1}{2} +\frac{1}{2}-\frac{1}{3} +...+\frac{1}{99} -\frac{1}{100} =1-\frac{1}{100}=1-0,01\\\\(a+0.01)^{2018}=(1-0.01+0.01)^{2018}= 1^{2018}=1$∈|N