a)
[tex]\it \dfrac{1}{n(n+2)}=\dfrac{a}{n}-\dfrac{b}{n+2}=\dfrac{an+2a-bn}{n(n+2)}=\dfrac{(a-b)n+2a}{n(n+2)} \Rightarrow \begin{cases} \ir a-b=0 \Rightarrow a=b\\ \\ \it 2a=1 \Rightarrow a=\dfrac{1}{2}\end{cases}\\ \\ \\ Deci:\ \ \dfrac{1}{n(n+2)}=\dfrac{\dfrac{1}{2}}{n}-\dfrac{\dfrac{1}{2}}{n+2}=\dfrac{1}{2}\Big(\dfrac{1}{n}-\dfrac{1}{n+2}\Big)[/tex]
b)
Folosim formula de la a) și expresia devine:
[tex]\it E=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\ ...\ +\dfrac{1}{96}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{100}\Big)=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{100}\Big)=\\ \\ \\ =\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}[/tex]
