Question

Let a sequence of number be as follows:

If $t_n$ is the first term of $n$ th row then $\underset{n\to \infty }{\mathrm{Lim}}\left(\sqrt{t_n}-n\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{-1}{2}$
• C. 1
• D. -1

Answer 1

Answer: (B)

$S_n=1+3+7+13+21+\dots +t_n$
$S_n=1+3+7+13+\dots +t_{n-1}+t_n$
Subtracting
$0=1+2+4+6+8+\dots +\left(t_n-t_{n-1}\right)-t_n$
$t_n=1+2+4+6+8+\dots +\left(t_n-t_{n-1}\right)$
$=1+\frac{n-1}{2}[2\times 2+(n-2)\times 2]$
$=1+(n-1)n=n^2-n+1$
$\underset{n\to \infty }{\lim }\left(\sqrt{n^2-n+1}-n\right)$
$=\underset{n\to \infty }{\lim }\frac{n^2-n+1-n^2}{\sqrt{n^2-n+1}+n}$
$=-\frac{1}{2}$