Question

Tính 

N = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/n.(n+1).(n+2)

Answer 1

\(N=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)

\(\Rightarrow2N=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)

\(\Rightarrow N=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right)\)

Answer 2

N=1/1.2.3 +1/2.3.4 +1/3.4.5 +...+1/n.(n+1).(n+2) 

⇒2N=2/1.2.3 +2/2.3.4 +2/3.4.5 +...+2/n.(n+1).(n+2) 

⇒N=1/2 .(1/1.2 −1/2.3 +1/2.3 −1/3.4 +1/3.4 −1/4.5 +...+1/n.(n+1) −1/(n+1).(n+2) )

⇒N=1/2 .(1/1.2 −1/(n+1).(n+2) )

chúc bạn học tốt !