Question

tính

\(D=\frac{2\cdot2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)

Answer 1

Ta có \(\frac{1}{1+2+3+..+n}=\frac{2}{n\left(n+1\right)}\)

Xét mẫu ta có

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2016}\)

\(=2\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2015\times2016}\right)\)

\(=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=2\left(1-\frac{1}{2017}\right)=\frac{2\times2016}{2017}\)

Thế vào ta được

\(D=\frac{2\times2016\times2017}{2\times2016}=2017\)

Answer 2

 = 2017