$k=1$
with $1$
cells. During day $k=2,3,\ldots$
the organism produces $\frac{k^2}{k1}$
times more new cells than it produced on day $k1$
. 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 $n.$

 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.
 Simplify
$\sum_{k=2}^n (\frac{1}{k}  \frac{1}{k1})$
.

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

Can you write your recurrence relationship as a nonrecursive 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$N_k=\frac{k^2}{k1}\,N_{k1},$
which we can expand as$$ =\frac{k^2}{k1}\cdot\frac{(k1)^2}{k2}\,N_{k2} \\ =\frac{k^2}{k1}\cdot\frac{(k1)^2}{k2}\cdot\frac{(k2)^2}{k3}\,N_{k3},$$
where we notice successive terms simplify, hence in the end we get:$$N_k = k^2 \cdot (k1) \cdots 1=k \cdot k!.$$
The total number of cells is thus
$$\sum_{k=1}^n k\cdot k!= \sum_{k=1}^n (k+11)k! = \sum_{k=1}^n (k+1)!  \sum_{k=1}^n k!,$$
where all the terms except the largest and smallest cancel each other leaving$(n+1)!1$
.