Question

A=1/1x2 + 1/2x3 + ... + 1/99x100

    Tính A

Answer 1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{100-99}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{99}{100}\)

Answer 2

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Answer 3

Theo đề bài ta có:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A=1-\frac{1}{100}\)

\(\Leftrightarrow A=\frac{99}{100}\)

Vậy \(\Rightarrow A=\frac{99}{100}\)

Answer 4

\(A=\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)