Avem urmatoarea formula:
[tex]\frac{m}{k(k+n)}=m\cdot\frac{n}{nk(k+n)}=\frac{m}{n}\cdot\frac{n+k-k}{k(k+n)}=\frac{m}{n}(\frac{k+n}{k(k+n)}-\frac{k}{k(k+n)})=\\=\boxed{\frac{m}{n}(\frac{1}{k}-\frac{1}{k+n})}[/tex]
[tex]S=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{15}{106\cdot121}\\\\
S=\frac{1}{1\cdot(1+1)}+\frac{2}{2\cdot(2+2)}+\frac{3}{4\cdot(4+3)}+\frac{4}{7\cdot(7+4)}+...+\frac{15}{106\cdot(106+15)}\\\\
S=\frac{1}{1}(\frac{1}{1}-\frac{1}{2})+\frac{2}{2}(\frac{1}{2}-\frac{1}{4})+\frac{3}{3}(\frac{1}{4}-\frac{1}{7})+..+\frac{15}{15}(\frac{1}{106}-\frac{1}{121})\\\\
[/tex]
[tex]S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{106}+\frac{1}{106}-\frac{1}{121}\\\\
\text{Se vor reduce toti termenii, mai putin primul si ultimul}\\\\
S=\frac{1}{1}-\frac{1}{121}=\boxed{\frac{120}{121}}[/tex]
