Question

Salut,am nevoie de ajutor la ex. 2.664C

Answer 1

$1+(1+2)+(1+2+3)+...+(1+2+...+n) = \\ \\ =\displaystyle \sum\limits_{k=1}^n\Big(\sum\limits_{i=1}^ki\Big) = \sum\limits_{k=1}^n\dfrac{k(k+1)}{2} =\sum\limits_{k=1}^n\dfrac{k^2+k}{2} = \\ \\ = \dfrac{1}{2} \sum\limits_{k=1}^n k^2+\dfrac{1}{2}\sum\limits_{k=1}^nk = \dfrac{n(n+1)(2n+1)}{12}+\dfrac{n(n+1)}{4}\\ \\ \\\text{Ne intereseaza doar coeficientul lui }n^3\text{ care este }\dfrac{2}{12}=\dfrac{1}{6}$

$\Rightarrow \lim\limits_{n\to \infty} \dfrac{kn^3+1+(1+2)+...+(1+2+...+n)}{n^3+n^2+1}=\dfrac{7}{6}\\\\ \Rightarrow\dfrac{k+\dfrac{1}{6}}{1} = \dfrac{7}{6} \Rightarrow \dfrac{6k+1}{6} = \dfrac{7}{6} \Rightarrow \boxed{k = 1}$

=> c) corect.