Question

Se aplică suma telescopică pentru:
2/(2*3)+2/(3*4)+......+2/(2018*2019)

Answer 1

$\displaystyle \sum\limits_{k=2}^{2018}\dfrac{2}{k(k+1)} = 2\sum\limits_{k=2}^{2018}\dfrac{1}{k(k+1)} = 2\sum\limits_{k=2}^{2018}\dfrac{(k+1)-k}{k(k+1)}= \\ \\\\ =2\sum\limits_{k=2}^{2018}\Bigg[\dfrac{k+1}{k(k+1)} -\dfrac{k}{k(k+1)}\Bigg] = 2\sum\limits_{k=2}^{2018}\Big(\dfrac{1}{k}-\dfrac{1}{k+1}\Big) =$

$\\ \\ \displaystyle=2\Bigg(\sum\limits_{k=2}^{2018}\dfrac{1}{k}-\sum\limits_{k=2}^{2018}\dfrac{1}{k+1}\Bigg) =\\ \\ \\ =2\Bigg[\dfrac{1}{2}+\sum\limits_{k=2}^{2017}\dfrac{1}{k+1} - \Bigg(\sum\limits_{k=2}^{2017}\dfrac{1}{k+1}+\dfrac{1}{2018+1}\Bigg)\Bigg] = \\ \\ \\ = 2\Big(\dfrac{1}{2}-\dfrac{1}{2019}\Big) = \dfrac{2(2019-2)}{2\cdot 2019} = \boxed{\dfrac{2017}{2019}}$

Answer 2

O suma de tipul  S=(a1-a2)+(a2-a3)+(a3-a4)+...+(an-an+1) se numeste suma telescopica.