An organism is born on day $k=1$ with $1$ cells. During day $k=2,3,\ldots$ the organism produces $\frac{k^2}{k-1}$ times more new cells than it produced on day $k-1$. Give a simplified expression for the total of all its cells after $n$ days. Hint: This is different to the number of new cells produced during day $n.$
1. Every day, a builder lays 2 more bricks than the total amount of bricks laid in the last 2 days. Express the number of bricks laid a day as a recursive formula.
2. Simplify $\sum_{k=2}^n (\frac{1}{k} - \frac{1}{k-1})$.

Try to formulate the number of new cells each day as a recurrence.

Can you write your recurrence relationship as a non-recursive expression?

The total number of cells on any day is the sum of the number of cells produced by the organism up to that day.

Rearrange $k\cdot k!$ into a difference between 2 factorials.

Denote with $N_k$ the number of cells grown at step $k$. We have the recurrence $N_k=\frac{k^2}{k-1}\,N_{k-1},$ which we can expand as $$=\frac{k^2}{k-1}\cdot\frac{(k-1)^2}{k-2}\,N_{k-2} \\ =\frac{k^2}{k-1}\cdot\frac{(k-1)^2}{k-2}\cdot\frac{(k-2)^2}{k-3}\,N_{k-3},$$ where we notice successive terms simplify, hence in the end we get: $$N_k = k^2 \cdot (k-1) \cdots 1=k \cdot k!.$$

The total number of cells is thus $$\sum_{k=1}^n k\cdot k!= \sum_{k=1}^n (k+1-1)k! = \sum_{k=1}^n (k+1)! - \sum_{k=1}^n k!,$$ where all the terms except the largest and smallest cancel each other leaving $(n+1)!-1$.