Question

Let a sequence of number be as follows:

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

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

Answer 1

Answer: (B)

$S_{n}=1+3+7+13+21+\ldots+t_{n}$
$S_{n}= 1+3+7+13+\ldots+t_{n-1}+t_{n}$
Subtracting
$0=1+2+4+6+8+\ldots+\left(t_{n}-t_{n-1}\right)-t_{n}$
$t_{n}=1+2+4+6+8+\ldots+\left(t_{n}-t_{n-1}\right)$
$=1+\frac{n-1}{2}[2 \times 2+(n-2) \times 2]$
$=1+(n-1) n=n^{2}-n+1$
$ \displaystyle \lim_{n\to\infty}\left(\sqrt{n^{2}-n+1}-n\right)$
$=\displaystyle \lim_{n\to\infty} \frac{n^{2}-n+1-n^{2}}{\sqrt{n^{2}-n+1}+n}$
$=-\frac{1}{2}$