Common Cases of Proof by Induction (Edexcel A Level Further Maths: Core Pure)

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Proof by Induction - Sequences

What are the steps for proof by induction with sequences?

  • STEP 1: The basic step
    • Show the result is true for the base case
      • If the recursive relation formula for the next term involves the previous two terms then you need to show the position-to-term formula works the first two given terms which will be given as part of the definition of the sequence
    • This is normally n = 1 or 0 but it could be any integer
      • For example: To prove u subscript n equals 3 to the power of n minus 2 is the position-to-term formula for the recursive sequence u subscript n plus 1 end subscript equals 3 u subscript n plus 4 comma space u subscript 1 equals 1
      • u subscript 1 equals 3 to the power of 1 minus 2 equals 1
  • STEP 2: The assumption step
    • Assume the result is true for n = k for some integer k
      • For example: Assume u subscript k equals 3 to the power of k minus 2 is true
    • There is nothing to do for this step apart from writing down the assumption
  • STEP 3: The inductive step
    • Using the assumption show the result is true for n = k + 1
    • It can be helpful to write down what you want to show
    • The assumption from STEP 2 will be needed at some point
      • For example:  Want to show u subscript k plus 1 end subscript equals 3 to the power of k plus 1 end exponent minus 2
    • The trick to this step is to use the recursive relation formula:
      • u subscript k plus 1 end subscript equals 3 u subscript k plus 4
  • STEP 4: The conclusion step
    • State the result is true
    • Explain in words why the result is true
    • It must include:
      • If true for n = k then it is true for n = k + 1
      • Since true for n = 1 the statement is true for all n ∈ ℤ, n ≥ 1  by mathematical induction
    • The sentence will be the same for each proof just change the base case from n = 1 if necessary

Worked example

A sequence is defined by u subscript n plus 1 end subscript equals 5 u subscript n minus 4 comma space u subscript 1 equals 2 for n greater or equal than 1.

Prove by mathematical induction that u subscript n equals 5 to the power of n minus 1 end exponent plus 1 for n greater or equal than 1.

9-1-2-edexcel-al-fm-cp-proof-by-induction-sequences-we

Proof by Induction - Series

What are the steps for proof by induction with series?

  • STEP 1: The basic step
    • Show the result is true for the base case
    • This is normally n = 1 or 0 but it could be any integer
      • For example: To prove sum from r equals 1 to n of r squared equals 1 over 6 n left parenthesis n plus 1 right parenthesis left parenthesis 2 n plus 1 right parenthesis is true for all integers n ≥ 1 you would first need to show it is true for n = 1: 
      • sum from r equals 1 to 1 of r squared equals 1 over 6 left parenthesis 1 right parenthesis left parenthesis left parenthesis 1 right parenthesis plus 1 right parenthesis left parenthesis 2 left parenthesis 1 right parenthesis plus 1 right parenthesis
  • STEP 2: The assumption step
    • Assume the result is true for n = k for some integer k
      • For example: Assume sum from r equals 1 to k of r squared equals 1 over 6 k left parenthesis k plus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis is true
    • There is nothing to do for this step apart from writing down the assumption
  • STEP 3: The inductive step
    • Using the assumption show the result is true for n = k + 1
    • It can be helpful to simplify LHS & RHS separately and show they are identical
    • The assumption from STEP 2 will be needed at some point
      • For example: L H S equals sum from r equals 1 to k plus 1 of r squared and R H S equals 1 over 6 open parentheses k plus 1 close parentheses open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses open parentheses 2 open parentheses k plus 1 close parentheses plus 1 close parentheses
    • The trick to this step is to use:

sum from r equals 1 to k plus 1 of f open parentheses r close parentheses equals f open parentheses k plus 1 close parentheses plus sum from r equals 1 to k of f open parentheses r close parentheses

  • STEP 4: The conclusion step
    • State the result is true
    • Explain in words why the result is true
    • It must include:
      • If true for n = k then it is true for n = k + 1
      • Since true for n = 1 the statement is true for all n ∈ ℤ, n ≥ 1  by mathematical induction
    • The sentence will be the same for each proof just change the base case from n = 1 if necessary

Worked example

Prove by induction that sum from r equals 1 to n of r open parentheses r minus 3 close parentheses equals 1 third n open parentheses n minus 4 close parentheses open parentheses n plus 1 close parentheses for n element of straight integer numbers to the power of plus.

1-5-1-ib-aa-hl-proof-by-induction-we-solution

Proof by Induction - Divisibility

What are the steps for proof by induction with series?

  • STEP 1: The basic step
    • Show the result is true for the base case
    • This is normally n = 1 or 0 but it could be any integer
      • For example: To prove f left parenthesis n right parenthesis equals 4 to the power of n minus 1 is divisible by 3 for all integers n ≥ 1 you would first need to show it is true for n = 1: 
      • f left parenthesis 1 right parenthesis equals 4 minus 1 equals 3 equals 3 left parenthesis 1 right parenthesis
  • STEP 2: The assumption step
    • Assume the result is true for n = k for some integer k
      • For example: Assume f left parenthesis k right parenthesis equals 4 minus 1 is divisible by 3
    • There is nothing to do for this step apart from writing down the assumption
    • The trick for this step is to write the expression as a multiple of the number
      • For example: 4 to the power of k minus 1 equals 3 p for some integer p
  • STEP 3: The inductive step
    • Using the assumption show the result is true for n = k + 1
    • The assumption from STEP 2 will be needed at some point
      • For example: Show that f left parenthesis k plus 1 right parenthesis equals 4 to the power of k plus 1 end exponent minus 1 is divisible by 3
    • The trick to this step is to use:

a to the power of k plus 1 end exponent equals a cross times a to the power of k

      • For example: f left parenthesis k plus 1 right parenthesis equals 4 left parenthesis 4 to the power of k right parenthesis minus 1 equals 4 left parenthesis 1 plus 3 p right parenthesis minus 1 equals 3 left parenthesis p plus 1 right parenthesis
    • Another trick is to show that f left parenthesis k plus 1 right parenthesis minus f left parenthesis k right parenthesis is divisible by the number
    • Be careful with numbers like 6 as for these you can instead show it is divisible by both 3 and 2
  • STEP 4: The conclusion step
    • State the result is true
    • Explain in words why the result is true
    • It must include:
      • If true for n = k then it is true for n = k + 1
      • Since true for n = 1 the statement is true for all n ∈ ℤ, n ≥ 1  by mathematical induction
    • The sentence will be the same for each proof just change the base case from n = 1 if necessary

Worked example

Prove by induction that f left parenthesis n right parenthesis equals 6 to the power of n plus 13 to the power of n plus 1 end exponent is divisible by 7 for all integers n greater or equal than 0.

9-1-2-edexcel-al-fm-cp-proof-by-induction-divisibility-we

Proof by Induction - Matrices

What are the steps for proof by induction with matrices?

  • STEP 1: The basic step
    • Show the result is true for the base case
    • This is normally n = 1 but it could be any integer
      • For example: To prove bold italic M to the power of n equals open parentheses table row cell 2 to the power of n end cell 0 row cell 1 minus 2 to the power of n end cell 1 end table close parentheses where bold italic M equals open parentheses table row 2 0 row cell negative 1 end cell 1 end table close parentheses for n greater or equal than 1
      • open parentheses table row cell 2 to the power of 1 end cell 0 row cell 1 minus 2 to the power of 1 end cell 1 end table close parentheses equals open parentheses table row 2 0 row cell negative 1 end cell 1 end table close parentheses
  • STEP 2: The assumption step
    • Assume the result is true for n = k for some integer k
      • For example: Assume bold italic M to the power of k equals open parentheses table row cell 2 to the power of k end cell 0 row cell 1 minus 2 to the power of k end cell 1 end table close parentheses is true
    • There is nothing to do for this step apart from writing down the assumption
  • STEP 3: The inductive step
    • Using the assumption show the result is true for n = k + 1
    • It can be helpful to write down what you want to show
    • The assumption from STEP 2 will be needed at some point
      • For example:  Want to show bold italic M to the power of k equals open parentheses table row cell 2 to the power of k plus 1 end exponent end cell 0 row cell 1 minus 2 to the power of k plus 1 end exponent end cell 1 end table close parentheses
    • The trick to this step is to use write:
      • bold italic M to the power of k plus 1 end exponent equals bold italic M bold italic M to the power of bold k
      • The entries in the matrix can get messy so it can be clearer to write bold italic M to the power of k plus 1 end exponent equals open parentheses table row a b row c d end table close parentheses and then find expressions for a, b, c, d on separate lines
  • STEP 4: The conclusion step
    • State the result is true
    • Explain in words why the result is true
    • It must include:
      • If true for n = k then it is true for n = k + 1
      • Since true for n = 1 the statement is true for all n ∈ ℤ, n ≥ 1  by mathematical induction
    • The sentence will be the same for each proof just change the base case from n = 1 if necessary

Worked example

The matrix bold italic M is given by M equals open parentheses table row 2 2 row 0 1 end table close parentheses.

Prove, using mathematical induction, that bold italic M to the power of n equals open parentheses table row cell 2 to the power of n end cell cell 2 left parenthesis 2 to the power of n minus 1 right parenthesis end cell row 0 1 end table close parentheses.

9-1-2-edexcel-al-fm-cp-proof-by-induction-matrices-we

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Dan

Author: Dan

Expertise: Maths

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.