\[\sum_{n=1}^{n}n^5=\frac{(n^2)(2n^2+2n-1)(n+1)^2}{12}\]
Boundary conditions - n is a natural number
Example For n=3 \[\sum_{n=1}^{3}n^5=1^5+2^5+3^5=\frac{(3^2)(2*3^2+2*3-1)(3+1)^2}{12}=276\]
For n=5 \[\sum_{n=1}^{5}n^5=1^5+2^5+3^5+4^5+5^5=\frac{(5^2)(2*5^2+2*5-1)(5+1)^2}{12}=4425\]
