Question

[Nguyen Duc Anh]

tính giá trị biểu thức:

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2009}\)

Answer 1

[alibaba nguyễn]

Áp dụng công thức: 

\(1+2+...+n=\frac{n\left(n+1\right)}{2}\) thì được

\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2009}\)

\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2009.2010}{2}}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2009.2010}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{1004}{1005}\)

Answer 2

[Thanh Tùng DZ]

thôi, làm luôn nè

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2009}\)

\(=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2009\right).2009:2}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2009.2010}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+2.\left(\frac{1}{4}-\frac{1}{5}\right)+...+2.\left(\frac{1}{2009}-\frac{1}{2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2.\frac{502}{1005}\)

\(=\frac{1004}{1005}\)