Question

Answer 1

\(a,A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\\ A=1-\dfrac{1}{n+1}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\\ b,2B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\\ 2B=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ 2B=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n^2+3n+2-2}{2\left(n+1\right)\left(n+2\right)}\\ B=\dfrac{2\left(n^2+3n\right)}{2\left(n+1\right)\left(n+2\right)}=\dfrac{n^2+3n}{n^2+3n+2}\)