Question

a) Chứng tỏ rằng với n\(\in\)N,n\(\ne\)0 thì:

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

b) áp dụng kết quả ở câu a) đẻ tính nhanh :

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

Answer 1

a)\(\Leftrightarrow\frac{1}{n\left(n+1\right)}=\frac{n+1-1}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)(đpcm)

b)\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{8}\)

\(\Rightarrow A=\frac{3}{8}\)