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Find the first n-term sum Tₙ of bₙ.

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

TOPIC 01

2023 JM01

The first reciprocal term is 1/6, not 1/2.

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01 / Standard#Your turn

Find the first n-term sum Tₙ of bₙ.

an=2⋅3n,bn=1an+log⁡2ana_n=2\cdot3^n,\quad b_n=\frac1{a_n}+\log_2a_n

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

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate reciprocal and logarithmic contributions.
Hint 2
Use a geometric sum and 1+⋯+n.
Worked solution
  1. Expand the logarithm.

    bn=12⋅3n+1+nlog⁡23b_n=\frac1{2\cdot3^n}+1+n\log_2 3
  2. Sum each component.

    Tn=14(1−3−n)+n+n(n+1)2log⁡23T_n=\frac14(1-3^{-n})+n+\frac{n(n+1)}2\log_2 3

Tₙ=(1−3⁻ⁿ)/4+n+n(n+1)log₂3/2.

Checks and common pitfalls: The first reciprocal term is 1/6, not 1/2.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Expand the logarithm.
    bn=12⋅3n+1+nlog⁡23b_n=\frac1{2\cdot3^n}+1+n\log_2 3
  • Sum each component.
    Tn=14(1−3−n)+n+n(n+1)2log⁡23T_n=\frac14(1-3^{-n})+n+\frac{n(n+1)}2\log_2 3

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Curriculum and source notes ↗