Question

Tính tổng sau :

1 /1x2 + 1/2x3 + 1/3x4 + ...+ 1/999x1000 + 1

Answer 1

Đặt A , ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(A=2-\frac{1}{1000}\)

\(A=\frac{2000}{1000}-\frac{1}{1000}\)

\(A=\frac{1999}{1000}\)

Answer 2

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

\(A=1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

Vậy \(A=\frac{1999}{1000}\)

Answer 3

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=\left(1-\frac{1}{1000}\right)+1\)

\(=\left(1-1\right)+\frac{1}{1000}\)

\(=\frac{1}{1000}\)