Sum of recursive sequence An organism is born on day $k=1$ with $1$ cells. During day $k = 2,3, \dots$ the organism produces $\frac{k^2}{k-1}$ times more cells than it had after day $k-1$. Find a simplified expression for the number of cells after $n$ days.
Denoting the total number of cells at the end of day $n$ by $C_n$, I found the recurrence $C_{n} = C_{n-1} + \frac{n^2}{n-1}C_{n-1}$ (for $n > 1$). But I have no idea how to go about solving this. The question is from a UK university admissions test which in theory does not require any specific knowledge about this type of expression.
(Edit: fixed recurrence)
 A: Lets start from day n. The organism will produce $$\frac{n^2}{n-1}$$ times as many cells as on day n-1, which will produce $$\frac{(n-1)^2}{n-2}$$ times as many cells as day n-2, and so on. This continues until day 1. This will be: $$(\frac{n^2}{n-1})(\frac{(n-1)^2}{n-2})(\frac{(n-2)^2}{n-3})....(\frac{3^2}{2})(\frac{2^2}{1})(1)$$ From this you can see that the denominator of the last fraction multiplied cancels a power on the numerator of the next fraction multiplied, from left to right. This cancels the multiplication to: $$n^2(n-1)(n-2)(n-3)....(3)(2)$$ This can be further reduced to: $$(n)(n)!$$ For day n-1 the total number produced will be $$(n-1)(n-1)!$$
You now need to add them all up. Let T be the total number of cells produced on the stated day. 
So $$(n^2)(T_{n-1}) + (n-1)^2(T_{n-2}) + (n-2)^2(T_{n-3}) + (n-3)^2(T_{n-4}).... + 4 + 1$$ Which is the same as: $$(n)(n)! + (n-1)(n-1)! + (n-2)(n-2)! + (n-3)(n-3)!..... + (3)(3)! + (2)(2)! + 1$$ This can be simplified to: $$\sum_{i=0}^{n-1} (n-i)(n-i)!$$Which is the same as: $$\sum_{i=1}^{n} (i)(i)!$$
Hope this helps.
