For the function given below, find a formula for the Riemann sum obtained by dividing the interval $[0,5]$ into $n$ equal subintervals and using the right-hand endpoint for each $c_k$. Then take a limit of this sum as $n\to \mathrm{\infty }$ to calculate the area under the curve over $[0,5]$.
$$f(x)=x^2+5$$
Write a formula for a Riemann sum for the function $f(x)=x^2+5$ over the interval $[0,5]$. $S_n=◻$ (Type an expression using $n$ as the variable.)

Step 1 ：First, we need to understand what a Riemann sum is. A Riemann sum is a certain kind of approximation of an integral by a finite sum. It is named after nineteenth century German mathematician Bernhard Riemann. One very common application is approximating the area of functions or lines on a graph, but also the length of curves and other approximations.

Step 2 ：For the function $f(x)=x^2+5$ over the interval $[0,5]$, we divide the interval into $n$ equal subintervals. The width of each subinterval is $\frac{5}{n}$.

Step 3 ：We use the right-hand endpoint for each $c_k$. So, $c_k=\frac{5k}{n}$ for $k=1,2,...,n$.

Step 4 ：Then, the Riemann sum is given by $S_n=\sum _{k=1}^{n}f(c_k)\mathrm{\Delta }x$, where $\mathrm{\Delta }x$ is the width of the subinterval.

Step 5 ：Substitute $f(c_k)$ and $\mathrm{\Delta }x$ into the formula, we get $S_n=\sum _{k=1}^n({\left(\frac{5k}{n}\right)}^2+5)\cdot \frac{5}{n}$.

Step 6 ：Simplify the formula, we get $S_n=\frac{125}{n^3}\sum _{k=1}^n{k}^2+\frac{25}{n}\sum _{k=1}^{n}1$.

Step 7 ：We know that $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum _{k=1}^{n}1=n$. Substitute these into the formula, we get $S_n=\frac{125}{6}(\frac{n(n+1)(2n+1)}{n^3})+25$.

Step 8 ：Simplify the formula, we get $S_n=\frac{125}{6}(2+\frac{3}{n}+\frac{1}{n^2})+25$.

Step 9 ：Take the limit of this sum as $n\to \mathrm{\infty }$, we get $\underset{n\to \mathrm{\infty }}{lim}{S}_n=\frac{125}{6}\cdot 2+25$.

Step 10 ：Calculate the limit, we get $\boxed{{\displaystyle \frac{275}{3}}}$.