Vom folosi urmatoarea formula:
[tex]\frac{1}{k(k+1)}=\frac{1+k-k}{k(k+1)}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)}=\boxed{\frac{1}{k}-\frac{1}{k+1}}[/tex]
Regula sirului este urmatoarea:
[tex]S=\frac{201}{1\cdot2}+\frac{601}{2\cdot3}+\frac{1201}{3\cdto4}+...+\frac{9001}{9\cdot10}\\
S=\frac{100(1\cdot2)+1}{1\cdot2}+\frac{100(2\cdot3)+1}{2\cdot3}+\frac{100(3\cdot4)+1}{3\cdot4}+...+\frac{100(9\cdot10)+1}{9\cdot10}\\
S=\underbrace{100+100+100+...+100}_{9\ \text{elemente}}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\\\\
[/tex]
[tex]S=900+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\\\\
\text{Se observa ca se reduc toti termenii, mai putin primul si ultimul}\\\\
S=900+\frac{1}{1}-\frac{1}{10}=901-\frac{1}{10}=\frac{9009}{10}[/tex]
