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# 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 $$n.$$

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## Warm-up Questions

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

## Hints

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