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For positive integer n, find the n maximising −5cₙ²+cₙ, where cₙ=2/aₙ and aₙ=2·3ⁿ.

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TOPIC 01

2023 JM01

The continuous optimum need not be an available sequence value.

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01 / Standard#Your turn

For positive integer n, find the n maximising −5cₙ²+cₙ, where cₙ=2/aₙ and aₙ=2·3ⁿ.

Official paper · jm01-2023 · II.3(c) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Complete the square in cₙ.
Hint 2
The available values are discrete powers 3⁻ⁿ.
Worked solution
  1. The continuous maximum is at c=1/10.

    −5c2+c=1/20−5(c−1/10)2-5c^2+c=1/20-5(c-1/10)^2
  2. Of the decreasing values 1/3,1/9,1/27,…, 1/9 is closest to 1/10.

    ∣1/9−1/10∣=1/90<1/10−1/27,n=2, f(2)=4/81|1/9-1/10|=1/90<1/10-1/27,\quad n=2,\ f(2)=4/81

n=2; the maximum is 4/81.

Checks and common pitfalls: The continuous optimum need not be an available sequence value.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The continuous maximum is at c=1/10.
    −5c2+c=1/20−5(c−1/10)2-5c^2+c=1/20-5(c-1/10)^2
  • Of the decreasing values 1/3,1/9,1/27,…, 1/9 is closest to 1/10.
    ∣1/9−1/10∣=1/90<1/10−1/27,n=2, f(2)=4/81|1/9-1/10|=1/90<1/10-1/27,\quad n=2,\ f(2)=4/81

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Curriculum and source notes ↗