Expected Cell Population

Consider a day $d$ when each cell from previous day can divide into $r$ new daughter cells where $1 \leq r \leq d$ .
As it's given that probability of cell $p(r)$ dividing into $r$ daughter cells is directly proportional to $r$ itself, this means $p(r) \propto r$ $p(r) = kr$ As, total probability must be equal to 1, we can say that $\sum _{r=1}^{d}p(r) = \sum_{r=1}^d kr = 1$ $k = \dfrac{2}{d(d+1)}$ $p(r) = \dfrac{2r}{d(d+1)}$ We know that each cell from day $d-1$ will behave exactly as before with the exception that it can now divide into atmost $d$ cells instead of $d-1$ . Hence, the expectation $E[d]$ equals $E[d] = E[d-1]\sum_{r=1}^d \dfrac{2r^2}{d(d+1)}$ We can also state $E[D]$ as $E[D] = \prod_{d=1}^D\sum_{r=1}^d \dfrac{2r^2}{d(d+1)} = \prod_{d=1}^D \dfrac{2}{d(d+1)}\dfrac{d(d+1)(2d+1)}{6} = \prod_{d=1}^D \dfrac{2d+1}{3}$ We can write $1\cdot3\cdot5\cdots(2n+1) = \dfrac{(2n+1)!}{2^n n!}$ So, $E[D] = \prod_{d=1}^D \dfrac{2d+1}{3} = \dfrac{(2D+1)!}{6^D D!}$ Putting $D = 14$ $E[14] = \dfrac{943496929375}{729}$ giving the answer as $\boxed{943496930104}$