[tex]x+1\ si \ x+2\ sunt \ numere\ consecutive\\
\overline{abc}=10\cdot 11=110\\
\overline{abc}=11\cdot 12=131\\
...................................................\\
\overline{abc}=31\cdot 32=992\\
S= \frac{1}{10\cdot 11} + \frac{1}{11\cdot 12} +...+ \frac{1}{31\cdot 32}\\
S= \frac{1}{10}- \frac{1}{11} + \frac{1}{11}- \frac{1}{12} +...+ \frac{1}{31}- \frac{1}{32} \\
S=\frac{1}{10}- \frac{1}{32}= \frac{11}{160} [/tex]
