Question

Dau coroană............

Answer 1

$S = \dfrac{2}{1\cdot 2\cdot 3}+\dfrac{2}{2\cdot 3\cdot 4}+...+\dfrac{2}{98\cdot 99\cdot 100} \\ \\ S = \sum\limits_{k=1}^{98}\dfrac{2}{k(k+1)(k+2)} = \sum\limits_{k=1}^{98}\dfrac{(k+2)-k}{k(k+1)(k+2)} =\\ \\ =\sum\limits_{k=1}^{98}\Big(\dfrac{k+2}{k(k+1)(k+2)}-\dfrac{k}{k(k+1)(k+2)}\Big) = \\ =\sum\limits_{k=1}^{98}\Big(\dfrac{1}{k(k+1)}- \dfrac{1}{(k+1)(k+2)}\Big) =$

$= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{98\cdot 99}-\dfrac{1}{2\cdot 3}-\dfrac{1}{3\cdot 4}-...-\dfrac{1}{99\cdot 100} = \\ \\ =\dfrac{1}{1\cdot 2}-\dfrac{1}{99\cdot 100} = \dfrac{1}{2}-\dfrac{1}{9900} = \dfrac{4950-1}{9900} = \dfrac{4949}{9900}$