Question

Ma poate ajuta și pe mine cineva cu exercițiul din imagine? Nu mai țin minte cum se fac... ​

Answer 1

 

$\displaystyle\\ \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots\frac{1}{97\cdot98}+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=?\\\\\text{Folosim formula:}~~~\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$\displaystyle\\\text{Rezolvare:}\\\\\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots\frac{1}{97\cdot98}+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=\\\\=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\\\\\text{Reducem termenii asemenea.}\\\\=\frac{1}{1}-\frac{1}{100}=\frac{100-1}{100}=\boxed{\bf\frac{99}{100}}$