16.11

To begin this problem, let us attempt to find the value of $\left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor$ for integers $n$ such that $0 \leq n \leq 2016$ .

Looking at the value $\frac{n^{2016}}{n^{2015}+1}$ , we see that this will be extremely close to $\frac{n^{2016}}{n^{2015}} = n$ because of the exponents which will lead to extremely large numbers for $n \geq 2$ . Seeing the additional $1$ that is added in the denominator of the fraction, we can deduce that the fraction $\frac{n^{2016}}{n^{2015}+1}$ will have a value slightly lower than $n$ for $n \geq 2$ . Therefore the greatest integer less than $\frac{n^{2016}}{n^{2015}+1}$ , or $\left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor$ , will be equivalent to $n-1$ for $n\geq 2$ .

So, we now know the value of $\left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor$ for the integers $n=2,3,4 \ldots 2014,2015,2016$ . Therefore, to find the value of $\sum_{n=0}^{2016} \left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor$ , we must simply add the values of $\left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor$ for $n=0,1$ to the value of $\sum_{n=2}^{2016} \left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor$ .

Therefore, $\sum_{n=0}^{2016} \left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor = \left\lfloor \frac{0^{2016}}{0^{2015}+1} \right\rfloor + \left\lfloor \frac{1^{2016}}{1^{2015}+1} \right\rfloor + \sum_{n=2}^{2016} \left\lfloor \frac{n^{2016}}{n^{2015}+1} \right\rfloor = \left\lfloor \frac{0}{1} \right\rfloor+ \left\lfloor \frac{1}{2} \right\rfloor + \sum_{n=2}^{2016} n-1 = 0 + 0 + 1 + 2 + 3 +\ldots + 2013 +2014 + 2015 =$

$1 + 2 + 3 +\ldots + 2013 +2014 + 2015 = \frac{(2015+1)\cdot2015}{2} = \frac{2016\cdot2015}{2} = 1008\cdot2015=\boxed{2031120}$