Question

Let $f\left(n\right)$ denote the $n^{th}$ term of the sequence $3,6,11,18,27,...$ and $g\left(n\right)$ denote the $n^{th}$ term of the sequence $3,7,13,21,...$ . Let $F\left(n\right)$ and $G\left(n\right)$ denote the sum of $n$ terms of the above sequences, respectiveley. $\underset{n\to \infty }{lim}\frac{F\left(n\right)}{G\left(n\right)}=$

• A. $2$
• B. $1$
• C. $0$
• D. $\infty$

Answer 1

The correct option is B $1$

$\underset{n\to \infty }{lim}=\frac{F\left(n\right)}{G\left(n\right)}f\left(n\right)=n^2+2,g\left(n\right)=n^2+n+1$
$F\left(n\right)=\sum n^2+2=\frac{n(n+1)(2n+1)}{6}+2n$
$=n(\frac{(n+1)(2n+1)}{6}+2)$
$=\frac{n}{6}(2n^2+3n+1+12)$
$=\frac{2n^3+3n^2+13n}{6}$

$G\left(n\right)=\sum (n^2+n+1)=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$
$=n[\frac{2n^2+3n+1}{6}+\frac{n+1}{2}+1]$
$=\frac{2n^3+6n^2+10n}{6}$

$\underset{n\to \infty }{lim}\frac{F\left(n\right)}{G\left(n\right)}=\underset{n\to \infty }{lim}\frac{2n^3+3n^2+13n}{2n^3+6n^2+10n}=1$