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. $0$
• B. $1$
• C. $2$
• D. $\infty$

Answer 1

The correct option is B $1$

$F\left(n\right)=\sum f\left(n\right);G\left(n\right)=\sum g\left(n\right)$
$S=3+6+11+18+27+....+t_{n-1}+t_n$
$-S=3+6+11+18+....+t_{n-2}+t_{n-1}+t_n$
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$0=3+(3+5+7+......+(t_{n-1}-t_{n-2})+(t_n-t_{n-1})-t_n)$
$t_n=3+\frac{n-1}{2}(6+(n-2\left)2\right)$
$t_n=3+(n^2-1)=n^2+2=f\left(n\right)$

$S=3+7+13+21+....+t_{n-1}+t_n$
$-S=3+7+13+....+t_{n-2}+t_{n-1}+t_n$
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$0=3+(4+6+8+......+(t_n-t_{n-1})-t_n$
$t_n=3+\frac{n-1}{2}(8+(n-2\left)2\right)$
$t_n=n^2+n+1=g\left(n\right)$
$\underset{n\to \infty }{lim}\frac{f\left(n\right)}{g\left(n\right)}=\frac{n^2+2}{n^2+n+1}=1$