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) - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 1

To find P(10), we substitute n = 10 into the formula:

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

Calculating the sums:

• ( \sum_{k=0}^{10} k^3 = 0^3 + 1^3 + 2^3 + ... + 10^3 = 0 + 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 + 729 + 1000 = 3025 )
• ( \sum_{k=0}^{9} k^3 = 0^3 + 1^3 + 2^3 + ... + 9^3 = 0 + 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 + 729 = 2025 )

Thus,

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

To solve the equation ( P(n) = 1.25 \times 10^8 ), we first express P(n) in terms of n:

$P(n) = \sum_{k=0}^{n} k^3 - \sum_{k=0}^{n-1} k^3 = n^3$

We set this equal to the given number:

$n^3 = 1.25 \times 10^8.$

Taking the cube root of both sides gives:

$n = \sqrt[3]{1.25 \times 10^8}.$

Calculating gives:

$n \approx 500.$