Gladly factored

$(1+x)^{2015} = \sum_{r=0}^{2015} x^r \dbinom{2015}{r}$ $\text{Differentiating with respect to x:}$ $2015(1+x)^{2014} = \sum_{r=0}^{2015} r x^{r-1} \dbinom{2015}{r}$ $\implies 2015x(1+x)^{2014} = \sum_{r=0}^{2015} r x^{r} \dbinom{2015}{r}$ $\text{Differentiating with respect to x:}$ $2015[2014x(1+x)^{2013} + (1+x)^{2014}] = \sum_{r=0}^{2015} r^2 x^{r-1} \dbinom{2015}{r}$ $\text{Substituting x =1:}$ $2015[2014\cdot2^{2013} + 2^{2014}] = \sum_{r=0}^{2015} r^2 \dbinom{2015}{r}$ $\therefore\sum_{r=0}^{2015} r^2 \dbinom{2015}{r} = (2^{2013})(2015)(2014+2) = (2^{2013})(2015)(2016)$ $2015 = 5\cdot13\cdot31$ $2016 = 2^5\cdot3^2\cdot7$ $\therefore\text{Required prime factors: 2, 3, 5, 7, 13, 31}$