Question

Answer 1

Bài 1:

a: \(n\cdot A=1+\dfrac{1}{n}+...+\dfrac{1}{n^{2021}}\)

=>\(A\cdot\left(n-1\right)=1-\dfrac{1}{n^{2022}}=\dfrac{n^{2022}-1}{n^{2022}}\)

=>\(A=\dfrac{\left(n^{2022}-1\right)}{n^{2022}\left(n-1\right)}\)

b: \(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

=2(1/2-1/3+1/3-1/4+...+1/99-1/100)

=2*49/100=49/50