Question

tìm n là số nguyên,biết:1+1/3+1/6+1/10+...+2/n.(n+1)=4039/2020 giúp mình câu này với

Answer 1

=>\(2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{4039}{2020}\)

=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{4039}{4040}\)

=>1-1/(n+1)=4039/4040

=>1/(n+1)=1/4040

=>n+1=4040

=>n=4039