Kenneth ChengMathematical modeling
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Concept of a sequence

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

高二選擇性必修 第二册(A版).pdf · 4.1 · PDF 7 / printed page 2

TOPIC 01

Concept of a sequence

Connect general terms, recurrence relations and partial sums.

What you will be able to explain

  • Connect general terms, recurrence relations and partial sums.
  • Justify the method and check the conditions in a new situation.

Defining relation

Connect general terms, recurrence relations and partial sums.

an=Sn−Sn−1 (n≥2)a_n=S_n-S_{n-1}\ (n\ge2)

Conditions

Indices are positive integers; a1=S1 needs separate treatment.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Does a partial-sum formula automatically work at n=0?

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

First six arithmetic terms: 2, 3, 4, 5, 6, 7; sum=27. The vertical display scale adjusts to include all six terms.

First six arithmetic terms: 2, 3, 4, 5, 6, 7; sum=27. The vertical display scale adjusts to include all six terms.

Explain: Calculate two valid cases and explain the change using the defining relation.

Transfer: Compare the first term with terms found by successive differences.

Try it. Leave your reasoning visible.

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

01 / Foundation#Worked example

Find a3.

an=2n+1a_n=2n+1
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
a3=3t+1a_3=3t+1
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    a3=2⋅3+1=7a_3=2\cdot3+1=7

  3. The index is substituted in every occurrence of n.

The requested value is 7.

Checks and common pitfalls: The index is substituted in every occurrence of n.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Find the smallest positive index with an>t.

an=2n−1,t=3a_n=2n-1,\quad t=3
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
2n−1>t2n-1>t
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    n>2n>2

  3. Apply the stated relation and retain its conditions.

    nmin⁡=3n_{\min}=3

  4. A strict inequality and an integer index require the next larger integer.

The requested value is 3.

Checks and common pitfalls: A strict inequality and an integer index require the next larger integer.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

Prove that the sequence is strictly increasing.

an=1−4n+1,n≥1a_n=1-\frac{4}{n+1},\quad n\ge1
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Compare consecutive terms.
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    an+1−an=4/[(n+1)(n+2)]>0a_{n+1}-a_n=4/[(n+1)(n+2)]>0

  3. Positive denominators and t>0 establish monotonicity for every allowed index.

The requested relation or conclusion is shown below.

an+1−an>0a_{n+1}-a_n>0

Checks and common pitfalls: Positive denominators and t>0 establish monotonicity for every allowed index.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find a3.

an=5n+1a_n=5n+1
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
a3=3t+1a_3=3t+1
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    a3=5⋅3+1=16a_3=5\cdot3+1=16

  3. The index is substituted in every occurrence of n.

The requested value is 16.

Checks and common pitfalls: The index is substituted in every occurrence of n.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Find a4 from the recurrence.

a1=2,an+1=an+6a_1=2,\quad a_{n+1}=a_n+6
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
a4=a1+3ta_4=a_1+3t
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    a2=8, a3=14, a4=20a_2=8,\ a_3=14,\ a_4=20

  3. Three transitions lead from index 1 to index 4.

The requested value is 20.

Checks and common pitfalls: Three transitions lead from index 1 to index 4.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Find a1.

Sn=n2+7(n≥1)S_n=n^2+7\quad(n\ge1)
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
a1=S1a_1=S_1
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    a1=12+7=8a_1=1^2+7=8

  3. The formula was not given at n=0; subtracting its extrapolated S0 would lose the constant.

The requested value is 8.

Checks and common pitfalls: The formula was not given at n=0; subtracting its extrapolated S0 would lose the constant.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Find a5.

Sn=8n2+1(n≥1)S_n=8n^2+1\quad(n\ge1)
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
a5=S5−S4a_5=S_5-S_4
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    a5=(200+1)−(128+1)=72a_5=(200+1)-(128+1)=72

  3. For n≥2 the constants cancel in successive differences.

The requested value is 72.

Checks and common pitfalls: For n≥2 the constants cancel in successive differences.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Find a5.

Sn=9n2+1(n≥1)S_n=9n^2+1\quad(n\ge1)
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
a5=S5−S4a_5=S_5-S_4
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    a5=(225+1)−(144+1)=81a_5=(225+1)-(144+1)=81

  3. For n≥2 the constants cancel in successive differences.

The requested value is 81.

Checks and common pitfalls: For n≥2 the constants cancel in successive differences.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Find the smallest positive index with an>t.

an=2n−1,t=10a_n=2n-1,\quad t=10
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
2n−1>t2n-1>t
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    n>5.5n>5.5

  3. Apply the stated relation and retain its conditions.

    nmin⁡=6n_{\min}=6

  4. A strict inequality and an integer index require the next larger integer.

The requested value is 6.

Checks and common pitfalls: A strict inequality and an integer index require the next larger integer.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Find the sum of the first t terms.

an=1n(n+1),t=11a_n=\frac1{n(n+1)},\quad t=11
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
1/[n(n+1)]=1/n−1/(n+1)1/[n(n+1)]=1/n-1/(n+1)
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    St=(1−1/2)+(1/2−1/3)+⋯+(1/11−1/12)S_t=(1-1/2)+(1/2-1/3)+\cdots+(1/11-1/12)

  3. Apply the stated relation and retain its conditions.

    St=1−1/12=11/12S_t=1-1/12=11/12

  4. Retain the two boundary terms after telescoping.

The requested value is 0.916666666667.

Checks and common pitfalls: Retain the two boundary terms after telescoping.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Prove that the sequence is strictly increasing.

an=1−12n+1,n≥1a_n=1-\frac{12}{n+1},\quad n\ge1
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Compare consecutive terms.
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    an+1−an=12/[(n+1)(n+2)]>0a_{n+1}-a_n=12/[(n+1)(n+2)]>0

  3. Positive denominators and t>0 establish monotonicity for every allowed index.

The requested relation or conclusion is shown below.

an+1−an>0a_{n+1}-a_n>0

Checks and common pitfalls: Positive denominators and t>0 establish monotonicity for every allowed index.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

Find the sum of the first t terms.

an=1n(n+1),t=13a_n=\frac1{n(n+1)},\quad t=13
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Use this intermediate relation.
1/[n(n+1)]=1/n−1/(n+1)1/[n(n+1)]=1/n-1/(n+1)
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    St=(1−1/2)+(1/2−1/3)+⋯+(1/13−1/14)S_t=(1-1/2)+(1/2-1/3)+\cdots+(1/13-1/14)

  3. Apply the stated relation and retain its conditions.

    St=1−1/14=13/14S_t=1-1/14=13/14

  4. Retain the two boundary terms after telescoping.

The requested value is 0.928571428571.

Checks and common pitfalls: Retain the two boundary terms after telescoping.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Prove that the sequence is strictly increasing.

an=1−14n+1,n≥1a_n=1-\frac{14}{n+1},\quad n\ge1
  • Indices are positive integers; a1=S1 needs separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the indexed terms or partial sums before simplifying.
Hint 2
Compare consecutive terms.
Worked solution

  1. Express the indexed terms or partial sums before simplifying.

  2. Apply the stated relation and retain its conditions.

    an+1−an=14/[(n+1)(n+2)]>0a_{n+1}-a_n=14/[(n+1)(n+2)]>0

  3. Positive denominators and t>0 establish monotonicity for every allowed index.

The requested relation or conclusion is shown below.

an+1−an>0a_{n+1}-a_n>0

Checks and common pitfalls: Positive denominators and t>0 establish monotonicity for every allowed index.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

Focus on one question

End-of-lesson check and correction

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    Teacher preparation and assessment

    Question sequence

    • Connect general terms, recurrence relations and partial sums.
    • Which condition is essential in concept of a sequence?
    • Does a partial-sum formula automatically work at n=0?

    Board plan

    • Defining relation: Connect general terms, recurrence relations and partial sums.
      an=Sn−Sn−1 (n≥2)a_n=S_n-S_{n-1}\ (n\ge2)
    • Conditions: Indices are positive integers; a1=S1 needs separate treatment.

    Anticipated thinking

    • Using S0 from a formula stated only for n≥1 can give a wrong first term.

    Assessment checklist

    • 1 mark: choose the correct representation and conditions.
    • 1 mark: establish the intermediate relation.
    • 1 mark: complete a connected calculation or proof.
    • 1 mark: interpret and check the conclusion.

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    Curriculum and source notes ↗