Question

O limite da soma lim (1+4+7+...+(3n+2))/(n+1)-(3n+1)/2 é:?
Ajuda com explicação pfv

Answer 1

Olá, Marlenguiliche.

$\lim\limits_{n\to+\infty}\frac{\overbrace{1+4+7+...+(3n-2)}^{\text{Soma da PA (1, ..., 3n-2), de raz\~ao 3, com n termos}}}{n+1}-\frac12(3n+1)=\\\\ =\lim\limits_{n\to+\infty}\frac{n\cdot\frac{1+3n-2}{2}}{n+1}-\frac{3n+1}2=\\\\ =\lim\limits_{n\to+\infty}\frac{n(3n-1)}{2(n+1)}-\frac{(3n+1)(n+1)}{2(n+1)}=\\\\ =\lim\limits_{n\to+\infty}\frac{3n^2-n-3n^2-3n-n-1}{2(n+1)}=\\\\ =\lim\limits_{n\to+\infty}\frac{-5n-1}{2n+2}=\\\\ =-\lim\limits_{n\to+\infty}\frac{n(5+\frac1 n)}{n(2+\frac2 n)}=$

$=-\lim\limits_{n\to+\infty}\frac{5+\frac1 n}{2+\frac2 n}=\\\\ =\boxed{-\frac52}$