Question

Matematica clasa a 11 a

Answer 1

$S_n = \sum\limits_{k=0}^{n} \dfrac{k^2+3k+1}{(k+2)!} = \sum\limits_{k=0}^{n}\dfrac{k^2+3k+2-1}{(k+2)!} = \\ \\ = \sum\limits_{k=0}^{n}\dfrac{(k+1)(k+2)-1}{(k+2)!} = \sum\limits_{k=0}^{n} \dfrac{(k+1)(k+2)}{(k+2)!} - \sum\limits_{k=0}^{n} \dfrac{1}{(k+2)!} = \\ \\ = \sum\limits_{k=0}^{n}\dfrac{(k+1)(k+2)}{k!(k+1)(k+2)} - \sum\limits_{k=0}^{n}\dfrac{1}{(k+2)!} =$

$\\ \\ = \sum\limits_{k=0}^{n} \dfrac{1}{k!} - \sum\limits_{k=0}^{n}\dfrac{1}{(k+2)!} = \\ \\ = \dfrac{1}{0!}+ \dfrac{1}{1!} + \dfrac{1}{2!} + ... + \dfrac{1}{n!} - \left[\dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + ... + \dfrac{1}{(n+2)!}\right] = \\ \\ = \dfrac{1}{0!} + \dfrac{1}{1!} + \left(\dfrac{1}{2!} + \dfrac{1}{3!}+... + \dfrac{1}{n!}\right) - \left( \dfrac{1}{2!} + \dfrac{1}{3!}+... + \dfrac{1}{n!}\right) - \\ \\ - \dfrac{1}{(n+1)!} - \dfrac{1}{(n+2)!} =$

$\\ \\ = \dfrac{1}{1} + \dfrac{1}{1} -\dfrac{1}{(n+1)!}- \dfrac{1}{(n+2)!}= \\ \\ = \boxed{2 -\left[\dfrac{1}{(n+1)!}+ \dfrac{1}{(n+2)!}\right] }$