[tex]\boxed{Forma\ generala:\frac{k}{n(n+k)}=\frac{n+k-n}{n(n+k)}=\frac{1}{n}-\frac{1}{n+k}}\\
Pt.\ k=1\Rightarrow \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\
pt. k=2\Rightarrow \frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}\\
Revenind:\\
a=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+_{\dots}+\frac{1}{99\cdot 100}\\
a=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+_{\dots}+\frac{1}{99}-\frac{1}{100}\\
a=1-\frac{1}{100}\Rightarrow \boxed{a=\frac{99}{100}}\\
\\
[/tex]
[tex]b=\frac{1}{2\cdot 4}+\frac{1}{4\cdot 6}+-{\dots}+\frac{1}{198\cdot 200}\\
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b=\frac{1}{2}(\frac{2}{2\cdot 4}+\frac{2}{4\cdot 6}+_{\dots}+\frac{2}{198\cdot 200}\\
\\
b=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+_{\dots}+\frac{1}{198}-\frac{1}{200})\\
\\
b=\frac{1}{2}(\frac{1}{2}-\frac{1}{200})\\
\\
b=\frac{1}{2}\cdot \frac{99}{200}\Rightarrow \boxed{b=\frac{99}{400}}\\[/tex]
