Question

Tìm x :

a) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

b) \(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

Answer 1

a) Đặt A=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{98\cdot99\cdot100}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+....+\frac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

2A=\(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\) =>A=\(\frac{4949}{9900}\div2=\frac{4949}{19800}\)

Đặt B=\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)

=>3B=\(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+....+\frac{3}{27\cdot28\cdot29\cdot30}\)

3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\)

3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{28\cdot29\cdot30}=\frac{1353}{8120}\)

=>B=\(\frac{1353}{8120}\div3=\frac{451}{8120}\)

Ta có : A-3x=B=>3x=A-B=\(\frac{4949}{19800}\)-\(\frac{451}{8120}\)\(\approx\frac{1}{5}\)=>x=\(\frac{1}{5}\div3\)=\(\frac{1}{15}\)