Question

Calculati limita sirului: 1*3+2*4+...+n(n+2) totul supra 1*2+2*3+...+n(n+1) ||| *=ori

Answer 1

$S_{1}=$1*2+2*3+...+n*(n+2)=∑n(n+2)=∑ n²+2∑n=$\frac{n(n+1)(2n+1)}{6}+ 2\frac{n(n+1)}{2}$=$\frac{n(n+1)(2n+7)}{6}$
∑n(n+1)=∑n²+∑n=$S_{2}= \frac{n(n+1)(2n+4)}{6}$
$\lim_{n \to \infty} \frac{ S_{1} }{ S_{2} }= \lim_{n \to \infty} \frac{n(n+1)(2n+7)}{n(n+1)(2n+4)}= \lim_{n \to \infty} \frac{2n+7}{2n+4}=1$