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 $n \rightarrow \infty$ 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 $[0,5]$. $S_{n}=\square$ (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 \(S_{n}=\sum_{k=1}^{n}f(c_{k})\Delta x\), where \(\Delta x\) is the width of the subinterval.

Step 5 ：Substitute \(f(c_{k})\) and \(\Delta x\) 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 \(n\rightarrow \infty\), we get \(\lim_{n\rightarrow \infty}S_{n}=\frac{125}{6}\cdot 2+25\).

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