Question

Tính tổng \(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

Answer 1

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

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

Vậy..

Answer 2

\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n+2-2}{4\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

Answer 3

2B=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+....+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{n\left(n+1\right)}\)-\(\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{2}\)-\(\dfrac{1}{n^2+3n+2}\)

=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)

=>B=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\):2

đến đây bạn tự tính nha !