Question

Ajutor urgeeent vă rog

Answer 1

$\it a)\ \dfrac{^{n+1)}1}{\ \ n}-\dfrac{^{n)}1}{\ n+1} = \dfrac{n+1-n}{n(n+1)}=\dfrac{1}{n(n+1)} \Longrightarrow \dfrac{1}{n(n+1)} =\dfrac{1}{n} -\dfrac{1}{n+1}\ \ \ \ (*) \\ \\ \\ b)\ N=\dfrac{1}{2\cdot3} +\dfrac{1}{3\cdot4} +\dfrac{1}{4\cdot5} +\ ...\ +\dfrac{1}{n(n+1)} \\ \\ \\ N\stackrel{(*)}{=}\ \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\ ...\ + \dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{n+1-2}{2(n+1)}=\\ \\ =\dfrac{n-1}{2n+2}$