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Calculate the next two terms of the recurrence.

Read the idea, work independently, then explain what changed.

TOPIC 01

2026 JM01

The recurrence iterates f; it does not substitute the index n into f.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Calculate the next two terms of the recurrence.

f(x)=3xx+3,x1=32,xn=f(xn−1)f(x)=\frac{3x}{x+3},\quad x_1=\frac32,\quad x_n=f(x_{n-1})

Official paper · jm01-2026 · II.5(a)(i) · PDF 5

Official original and suggested answers ↗ · Suggested answer PDF page 8

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Substitute x₁ into f.
Hint 2
Then use the newly obtained x₂.
Worked solution
  1. Evaluate the first substitution.

    x2=3(3/2)3/2+3=1x_2=\frac{3(3/2)}{3/2+3}=1
  2. Apply the same function once more.

    x3=3⋅11+3=34x_3=\frac{3\cdot1}{1+3}=\frac34

x₂=1, x₃=3/4.

Checks and common pitfalls: The recurrence iterates f; it does not substitute the index n into f.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Evaluate the first substitution.
    x2=3(3/2)3/2+3=1x_2=\frac{3(3/2)}{3/2+3}=1
  • Apply the same function once more.
    x3=3⋅11+3=34x_3=\frac{3\cdot1}{1+3}=\frac34

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Curriculum and source notes ↗