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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 Mechanics - Question 8 - 2019 - Paper 1
Question 8
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 \t... show full transcript
Worked Solution & Example Answer: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 Mechanics - Question 8 - 2019 - Paper 1
Step 1
Step 2
Find P(10)
Answer
To find P(10), we perform the same computation:
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Calculate ( \sum_{k=0}^{10} k^3 ):
- Using the formula ( \left( \frac{n(n+1)}{2} \right)^2 ), we find: [ \left( \frac{10(10+1)}{2} \right)^2 = \left( \frac{10 \times 11}{2} \right)^2 = (55)^2 = 3025 ]
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Calculate ( \sum_{k=0}^{9} k^3 ):
- Similarly, [ \left( \frac{9(9+1)}{2} \right)^2 = \left( \frac{9 \times 10}{2} \right)^2 = (45)^2 = 2025 ]
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Now, for P(10): [ P(10) = 3025 - 2025 = 1000 ]
Step 3
Solve the equation P(n) = 1.25 \times 10^8
Answer
To solve the equation, we know:
[ P(n) = \sum_{k=0}^{n} k^3 - \sum_{k=0}^{n-1} k^3 = n^3 ]
So we need to solve: [ n^3 = 1.25 \times 10^8 ]
Taking the cube root of both sides: [ n = \sqrt[3]{1.25 \times 10^8} ]
Calculating gives: [ n = 500 ]