P(n) = \sum_{k=0}^{n} k^3 - \sum_{k=0}^{n-1} k^3 \text{ where } n \text{ is a positive integer.} 8 (a) Find P(3) and P(10) 8 (b) Solve the equation P(n) = 1.25 \times 10^8 - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 1

To find P(10), we calculate:

$P(10) = \sum_{k=0}^{10} k^3 - \sum_{k=0}^{9} k^3$

Calculating each sum:

1. Using the formula for the sum of cubes: (\left( \frac{n(n + 1)}{2} \right)^2), for (n = 10):
   • (\sum_{k=0}^{10} k^3 = \left( \frac{10 \times 11}{2} \right)^2 = 55^2 = 3025)
2. For (n = 9):
   • (\sum_{k=0}^{9} k^3 = \left( \frac{9 \times 10}{2} \right)^2 = 45^2 = 2025)

Thus, we have:

$$P(10) = 3025 - 2025 = 1000.$$

To solve the equation, we set:

$P(n) = n^4\text{ (since } P(n) = (\frac{n(n + 1)}{2})^2\text{)} \Rightarrow n^4 = 1.25 \times 10^8$

Taking the fourth root:

$n = \sqrt[4]{1.25 \times 10^8} = \sqrt[4]{1.25} \times 10^2$

Calculating further:

1. (\sqrt[4]{1.25} \approx 1.118)
2. Therefore, (n \approx 1.118 \times 100 = 111.8)

Given that (n) must be a positive integer, we round this to: n = 500.