# Practice Paper 1 Question 11

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 \(n.\)

## Related topics

## Warm-up Questions

- 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 \(\sum_{k=2}^n (\frac{1}{k} - \frac{1}{k-1})\).

## Hints

- Hint 1Try to formulate the number of new cells each day as a recurrence.
- Hint 2Can you write your recurrence relationship as a non-recursive expression?
- Hint 3The total number of cells on any day is the sum of the number of cells produced by the organism up to that day.
- Hint 4Rearrange \(k\cdot k!\) into a difference between 2 factorials.

## Solution

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 \((n+1)!-1\).

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