Question

tính P=-1+1/2.1+1/3.2+1/4.3+..........+1/2018.2017+1/2018

Answer 1

\(P=\)\(-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)

\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)

\(P=-1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)

\(P=-1+1-\frac{1}{2018}+\frac{1}{2018}\)

\(P=0\)

Answer 2

\(P=-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)

\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)

\(P=-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)

P = 0