Question

Se considera numarul real N=1/6 + 1/2 + 1/20 +.....+ 1/n(n+1) a) Aratati ca 1/n(n+1)=1/n - 1/n+1 b) Determinati n astfel incat N=999/2000

Answer 1

$N = \sum\limits_{k=2}^{n}\dfrac{1}{k\cdot (k+1)} = \sum\limits_{k=2}^{n}\dfrac{k+1-k}{k\cdot (k+1)} = \\ \\ = \sum\limits_{k=2}^{n}\Big(\dfrac{k+1}{k\cdot (k+1)}-\dfrac{k}{k\cdot(k+1)}\Big) = \sum\limits_{k=2}^{n}\Big(\dfrac{1}{k}-\dfrac{1}{k+1}\Big) = \\ \\ = \sum\limits_{k=2}^n\dfrac{1}{k} - \sum\limits_{k=2}^n\dfrac{1}{k+1} =$

$=\Big(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\Big)-\Big(\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}+\dfrac{1}{n+1}\Big) = \\ \\ = \dfrac{1}{2}+\Big(\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\Big)-\Big(\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\Big) - \dfrac{1}{n+1} = \\ \\ = \dfrac{1}{2}-\dfrac{1}{n+1}$

$N = \dfrac{999}{2000} \Rightarrow \dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{999}{2000} \Rightarrow \dfrac{1}{n+1} = \dfrac{1}{2}-\dfrac{999}{2000} \Rightarrow \\ \\ \Rightarrow \dfrac{1}{n+1}=\dfrac{1000}{2\cdot 1000} - \dfrac{999}{2000} \Rightarrow \dfrac{1}{n+1} = \dfrac{1000-999}{2000} \Rightarrow \\ \\ \Rightarrow \dfrac{1}{n+1} = \dfrac{1}{1999+1} \Rightarrow \boxed{n = 1999}$