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An Explicit Rule Finds a Term Directly from Its Position

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Predicting What Comes Next: Exploring Sequences and Progressions · Explicit and recursive rules · 3:36

You'll learn: Substitute a positive integer index into a term formula.

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An Explicit Rule Finds a Term Directly from Its Position

3:36

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An Explicit Rule Finds a Term Directly from Its Position

You'll learn: Substitute a positive integer index into a term formula.

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Same topic

Explicit and recursive rules

7

Work Backwards to Find a Term’s Position
3:41
Work Backwards to Find a Term’s Position
You'll learn: Solve a term equation and test whether the index is a positive integer.
Generate Sequences from Linear and Quadratic Rules
3:50
Generate Sequences from Linear and Quadratic Rules
You'll learn: Evaluate several explicit rules and compare the resulting sequences.
A Recursive Rule Needs a Starting Value
3:40
A Recursive Rule Needs a Starting Value
You'll learn: Generate terms from the preceding term and compare with an explicit description.
Use a Recursive Machine That Doubles and Adds Three
3:43
Use a Recursive Machine That Doubles and Adds Three
You'll learn: Calculate successive terms and check a proposed value by continuing the rule.
A Recursive Rule Can Grow Much Faster Than Addition
3:41
A Recursive Rule Can Grow Much Faster Than Addition
You'll learn: Evaluate a rule that multiplies the previous term by one less than itself.
The Next Virahanka–Fibonacci Term Uses the Previous Two
3:46
The Next Virahanka–Fibonacci Term Uses the Previous Two
You'll learn: Generate the sequence from two initial values and explain why both are needed.
A Rule Can Use the Previous Three Terms
3:46
A Rule Can Use the Previous Three Terms
You'll learn: Apply a three-term recurrence and contrast its initial conditions with the two-term rule.

Next topic

Arithmetic progressions

5

Recognise an Arithmetic Progression by Its Constant Difference
3:50
Recognise an Arithmetic Progression by Its Constant Difference
You'll learn: Compare successive differences and identify the special class of sequences.
Why an AP’s nth Term Uses n−1 Steps
3:35
Why an AP’s nth Term Uses n−1 Steps
You'll learn: Count the jumps from the first term to derive tₙ=a+(n−1)d.
APs Can Decrease or Have Fractional Terms
3:43
APs Can Decrease or Have Fractional Terms
You'll learn: Identify a and d in positive, negative, fractional and decimal examples.
Plot an AP as Discrete Points on a Straight Line
3:56
Plot an AP as Discrete Points on a Straight Line
You'll learn: Turn a term table into ordered pairs(index, value).
Write the Same AP Explicitly and Recursively
3:48
Write the Same AP Explicitly and Recursively
You'll learn: Translate between tₙ=a+(n−1)d and tₙ=tₙ₋₁+d.

Next topic

AP applications and recovery problems

3

Find a Distant AP Term or Test Membership
3:28
Find a Distant AP Term or Test Membership
You'll learn: Use the general term forwards and backwards.
Recover an AP from Two Term Conditions
3:43
Recover an AP from Two Term Conditions
You'll learn: Use differences between term equations to find a and d.
Taxi Fares and Salaries Can Form APs
3:43
Taxi Fares and Salaries Can Form APs
You'll learn: Identify the first observed term and the fixed increment in a context.