Question

Chứng minh rằng: \(\frac{1}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)

Áp dụng tính : A=\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)

Answer 1

Ta co \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)

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

Ap dung cong thuc tren:

=> A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{132}\)

     A = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{11.12}\)

     A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{11}-\frac{1}{12}\)

     A = \(\frac{1}{2}-\frac{1}{12}\)

     A = \(\frac{5}{12}\)

Answer 2

Ta có: \(\frac{1}{a\left(a+1\right)}=\frac{\left(a+1\right)-a}{a\left(n+1\right)}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)

=> đpcm

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{2}-\frac{1}{12}\)

\(A=\frac{5}{12}\)