Question

If $f:R\to R$ is given by $f\left(x\right)=x+1,$ then the value of $\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+...+f\left(\frac{5(n-1)}{n}\right)],$ is:

• A. $\frac{7}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{7}{2}$

$f\left(0\right)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+....+f\left(\frac{5(n-1}{n}\right)$
$\Rightarrow 1+1+\frac{5}{n}+1+\frac{10}{n}+...+1+\frac{5(n-1)}{n}$
$\Rightarrow n+\frac{5}{n}\frac{(n-1)n}{2}=\frac{2n+5n-5}{2}=\frac{7n-5}{2}$
$\underset{n\to \infty }{lim}\frac{1}{n}\left(\frac{7n-5}{2}\right)=\frac{7}{2}$