jmc

To investigate this summation, we can start by looking at the first terms: $1+1+(2!)\times 2=2+(2!)\times 2=6=3!$. Thus, adding the next term, $6+(3!)\times 3=3!+(3!)\times 3=4\times 3!=24=4!$. Indeed, we see from the identity $$n!+(n!)\times n=(n+1)\times n!=(n+1)!$$that the sum will be equal to $51!$. The largest prime number to divide this quantity will be the largest prime number less than or equal to $51$, which is $\boxed{47}$.