Question

The value of $\underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $\frac{1}{6}$
• D. $6$

Answer 1

The correct option is C

$\frac{1}{6}$

Explanation for the correct option:

Given, $\underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]$

We know that the sum of $n$ natural numbers is $\frac{n(n+1)}{2}$.

Simplifying the given expression,

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]=\underset{n\to \infty }{\lim }\left[\frac{{\displaystyle \frac{n(n+1)}{2}}}{3n^2+5}\right] \\ \Rightarrow \underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]=\frac{1}{2}\underset{n\to \infty }{\lim }\left[\frac{{\displaystyle n^2+n}}{3n^2+5}\right] \\ \Rightarrow \underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]=\frac{1}{2}\underset{n\to \infty }{\lim }\left[\frac{{\displaystyle 1+\frac{1}{n}}}{3+{\displaystyle \frac{5}{n^2}}}\right] \\ \Rightarrow \underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]=\frac{1}{2}\left[\frac{{\displaystyle 1}}{3}\right] \\ \Rightarrow \underset{n\to \infty }{\lim }\left[\frac{(1+2+...+n)}{3n^2+5}\right]=\frac{1}{6} \end{gathered}$$

Hence, option C is the correct answer.