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Proof by mathematical induction. Introduction. Proof by mathematical induction is an extremely powerful tool for proving mathematical statements As we know, proof is essential in Maths as although something may seem to work for a number of cases, we need to be sure it will work in every case - PowerPoint PPT Presentation
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Proof by mathematicalinduction
Introduction• Proof by mathematical induction is an extremely
powerful tool for proving mathematical statements
• As we know, proof is essential in Maths as although something may seem to work for a number of cases, we need to be sure it will work in every case
• You have seen some of the formulae used in the series chapter – the had to be proven to work for every case before mathematicians could confidently use them
Teachings for Exercise 6A
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
The way ‘proof by mathematical induction’ works is often likened to
knocking dominoes over
If the dominoes are lined up, then you knock over the first one, every
domino afterwards will fall downiPhone Dominoes
Mathematically, if we want to prove that something is true for all possible
cases, we cannot do it numerically (as the numbers would just go on
forever)
However, if we show that if one case is true, and so is the next case, then we can therefore show it is true for
every case…
6A
How this works mathematically
BASIS Show that the statement to be proven works for the case n = 1
ASSUMPTION Assume that the statement is true for n = k (just replace the ns with ks!)
INDUCTIVE Show that if the statement is true for n = k, it is also true for n = k + 1 (ie – the next case)
This is harder to explain without an example. Essentially you find a way to express the next ‘case’ using k and show that it is equivalent to replacing k with ‘k + 1’
CONCLUSION You have shown that if the statement is true for one case, it must be true for the next
As it was true for n = 1, it must therefore be true for n = 2, 3, 4 and so on, PROVING the statement!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statement…
6A
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
This means ‘n can be any
positive integer’
This is the formula for the
sequence
This is the formula for the sum of the first n terms of the
sequence
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statement…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
BASIS Show that the statement is true for n = 1
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
∑𝑟=1
1
(2𝑟−1 )
¿1
(1)2
¿1
Replace n with 1 Replace n with 1
There will only be one term here, that we get by subbing n
= 1 into the expression
Calculate
The statement given is therefore true for n = 1
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statement…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
ASSUMPTION Assume the statement is true for n = k
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
∑𝑟=1
𝑘
(2𝑟−1 )=¿¿1+3+5+7+9+.. ..+(2𝑘−1)¿𝑘2
Write out the first few terms in the
sequence, and the last term, which will be in
terms of k
We are going to assume that this sequence is true for k, and hence the sum
will be equal to k2
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statement…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
ASSUMPTION Assume the statement is true for n = k
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
∑𝑟=1
𝑘
(2𝑟−1 )=¿¿1+3+5+7+9+.. ..+(2𝑘−1)¿𝑘2
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
∑𝑟=1
𝑘+1
(2𝑟−1 )=¿¿1+3+5+7+9+.. ..+(2𝑘−1 )¿1+3+5+7+9+ ....+(2𝑘−1 )+(2𝑘+1)
¿𝑘2+(2𝑘+1)
You can replace the first part as we assumed it was equal to k2 earlier
¿ (𝑘+1 )2
+(2 (𝑘+1 )−1)
The sequence will be the same, but with an extra term (sub in (k + 1)) for it!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statement…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
∑𝑟=1
𝑘
(2𝑟−1 )=¿𝑘2¿ ∑𝑟=1
𝑘+1
(2𝑟−1 )=¿(𝑘+1)2 ¿
We assumed that for n = k, the sum of the
series would be equal to k2
Using this assumption, we showed that the
summation for (k + 1) is equal to (k + 1)2
∑𝑟=1
𝑛
(2𝑟−1 )=𝑛2
So if the statement is true for one value, it will therefore be true for the next value
As it is true for the next value, it will therefore be true for the value after that, and so on…
However, this all relies on the assumption being correct…
Remember for the BASIS step, we showed that the statement is true for n = 1?
Well because it is true for n = 1, it must therefore be true for n = 2, n = 3……… and so on!
The statement is therefore true for all values of n!
CONCLUSION
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2)=1
6𝑛 (𝑛+1 ) (2𝑛+1 )
BASIS
∑𝑟=1
𝑛(𝑟 2)=1
6𝑛 (𝑛+1 ) (2𝑛+1 )
∑𝑟=1
1
(𝑟 2)
¿1
16 (1)(1+1 ) (2+1 )
¿1
Replace n with 1 Replace n with 1
There will only be one term
here, that we get by subbing n = 1
into the expression
Calculate
The statement given is therefore true for n = 1
Show that the statement is true for n = 1
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2)=1
6𝑛 (𝑛+1 ) (2𝑛+1 )
ASSUMPTION Assume the statement is true for n = k
∑𝑟=1
𝑘(𝑟 2)=¿ ¿1+4+9+16………+𝑘2¿ 16 𝑘 (𝑘+1 ) (2𝑘+1 )
Write out the first few terms in the
sequence, and the last term, which will be in
terms of k
We are going to assume that this sequence is true for k, and hence the sum
will be equal to the expression above
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2)=1
6𝑛 (𝑛+1 ) (2𝑛+1 )
ASSUMPTION Assume the statement is true for n = k
∑𝑟=1
𝑘(𝑟 2)=¿ ¿1+4+9+16………+𝑘2¿ 16 𝑘 (𝑘+1 ) (2𝑘+1 )
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
∑𝑟=1
𝑘+1
(𝑟 2)=¿ ¿1+4+9+16………+𝑘2+(𝑘+1)2
The sequence will be the same, but with an extra term (sub in (k + 1)) for it!
¿16 𝑘 (𝑘+1 ) (2𝑘+1 )+(𝑘+1)2
Replace the first part with the assumed formula from earlier!
This requires more simplification which will be shown on the next slide!!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2)=1
6𝑛 (𝑛+1 ) (2𝑛+1 )
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
∑𝑟=1
𝑘+1
(𝑟 2)=¿ ¿1+4+9+16………+𝑘2+(𝑘+1)2
The sequence will be the same, but with an extra term (sub in (k + 1)) for it!
¿16 𝑘 (𝑘+1 ) (2𝑘+1 )+(𝑘+1)2
Replace the first part with the assumed formula from earlier!
¿𝑘 (𝑘+1 ) (2𝑘+1 )
6+6 (𝑘+1 )2
6
¿𝑘 (𝑘+1 ) (2𝑘+1 )+6 (𝑘+1)2
6
¿(𝑘+1)𝑘(2𝑘+1)+6 (𝑘+1)¿ ¿6
¿(𝑘+1)2𝑘2+7𝑘+6¿ ¿6
¿ 6(𝑘+1)(𝑘+2)(2𝑘+3)
Rewrite both as fractions over 6
Combine
‘Clever factorisation’
method!Expand and simplify the inner brackets
Factorise the inner part
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2)=1
6𝑛 (𝑛+1 ) (2𝑛+1 )
CONCLUSION Explain why it proves the original statement
¿16 𝑘 (𝑘+1 ) (2𝑘+1 ) ¿
16 (𝑘+1) (𝑘+2 ) (2𝑘+3 )
¿16 (𝑘+1) (𝑘+1+1 ) (2(𝑘+1)+1 )
For n = k For n = (k + 1)
Rewrite some of the brackets
Written in this way, you can see that the k’s in the first statement have all been replaced with
‘k + 1’s So the statement was true for n = 1
We also showed that if it is true for one statement, it is true for the next
Therefore the formula has been proven!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]
BASIS
∑𝑟=1
𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]
∑𝑟=1
1
(𝑟 2𝑟 )
¿2
2 [1+(1−1)21 ]
¿2
Replace n with 1 Replace n with 1
There will only be one term
here, that we get by subbing n = 1
into the expression
Calculate
The statement given is therefore true for n = 1
Show that the statement is true for n = 1
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]
ASSUMPTION Assume the statement is true for n = k
∑𝑟=1
𝑛(𝑟 2𝑟 )=¿¿2+8+24+64……+𝑘(2𝑘)¿2 [1+(𝑘−1 )2𝑘 ]
Write out the first few terms in the
sequence, and the last term, which will be in
terms of k
We are going to assume that this sequence is true for k, and hence the sum
will be equal to the expression above
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]
ASSUMPTION Assume the statement is true for n = k
∑𝑟=1
𝑛(𝑟 2𝑟 )=¿¿2+8+24+64……+𝑘(2𝑘)¿2 [1+(𝑘−1 )2𝑘 ]
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term The sequence will be the same, but with an extra
term (sub in (k + 1)) for it!
∑𝑟=1
𝑛(𝑟 2𝑟 )=¿¿2+8+24+64……+𝑘(2𝑘)+(𝑘+1)2𝑘+1
Replace the first part with the assumed formula from earlier!
¿2 [1+(𝑘−1 )2𝑘 ]+(𝑘+1)2𝑘+1
¿2+2(𝑘−1)2𝑘+(𝑘+1)2𝑘+1
The simplification for this is difficult
You need to aim for the power of 2 to be ‘k + 1’ (as it was ‘k’ originally)
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
¿2+2(𝑘−1)2𝑘+(𝑘+1)2𝑘+1
2 x 2k = 2k+1
(add the powers)¿2+(𝑘−1)2𝑘+1+(𝑘+1)2𝑘+1
¿2+(𝑘−1+𝑘+1)2𝑘+1
In total, we have (k – 1) + (k + 1) 2k+1s
¿2+2𝑘2𝑘+1Simplify the bracket
¿2(1+𝑘2𝑘+1)Re-factorise the bracket
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
∑𝑟=1
𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]
CONCLUSION Explain why this shows the statement is true
2(1+𝑘2𝑘+1)2 [1+(𝑘−1 )2𝑘 ]
For n = k For n = (k + 1)
2(1+(𝑘+1−1)2𝑘+1)
Rewrite the first ‘k’ as ‘k + 1 – 1’
Written in this way, you can see that the k’s in the first statement have all been replaced with
‘k + 1’s So the statement was true for n = 1
We also showed that if it is true for one statement, it is true for the next
Therefore the formula has been proven!
You will need to become familiar with manipulating powers in the way shown
here!
Teachings for Exercise 6B
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers
You follow the same steps as before!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
BASIS Show that the statement is true for n = 1𝑓 (𝑛)=32𝑛+11
𝑓 (1 )=32(1)+11
¿20
Sub in n = 1
Calculate
20 is divisible by 4, so the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers
You follow the same steps as before!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
ASSUMPTION Assume the statement is true for n = k
𝑓 (𝑘 )=32𝑘+11𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒𝑏𝑦 4 𝑓𝑜𝑟 𝑘∈ℤ+¿ ¿
INDUCTIVE Show that the statement is then true for n = (k + 1)
𝑓 (𝑘+1 )=32(𝑘+ 1)+11
𝑓 (𝑘+1 )=32𝑘+2+11
𝑓 (𝑘+1 )=32𝑘×32+11
𝑓 (𝑘+1 )=9(3¿¿2𝑘)+11¿
Multiply out the bracket
32k+2 = 32k x 32 (adding powers when multiplying)So we have 9
lots of 32k
At this point we will combine the expressions for f(k) and f(k + 1) in order to prove that the statement is always divisible
by 4
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers
You follow the same steps as before!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
𝑓 (𝑘 )=32𝑘+11
INDUCTIVE Show that the statement is then true for n = (k + 1)
𝑓 (𝑘+1 )=9(3¿¿2𝑘)+11¿
𝑓 (𝑘+1 )− 𝑓 (𝑘)=¿[9 (32𝑘 )+11]− [32𝑘−11 ]
𝑓 (𝑘+1 )− 𝑓 (𝑘)=¿8 (32𝑘)
𝑓 (𝑘+1 )− 𝑓 (𝑘)=¿4 [2(32𝑘) ]
𝑓 (𝑘+1 )=¿𝑓 (𝑘 )+4 [2(32𝑘) ]
Subtract f(k) from f(k + 1), using the expressions above
Group terms on the right
sideTake out 4 as
a factor
Add f(k)
This shows that f(k + 1) is just f(k) with an expression added on
We assumed f(k) was divisible by 4 The expression to be added is divisible by 4
So the answer must be divisible by 4, if f(k) is! As the first case (n = 1) was divisible by 4, the
statement must be true!
CONCLUSION
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible
by 3 for all positive integers
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
BASIS Show that the statement is true for n = 1𝑓 (𝑛)=𝑛3−7𝑛+9
Sub in n = 1𝑓 (1 )=(1)3−7(1)+9
¿3Calculate
3 is divisible by 3, so the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible
by 3 for all positive integers
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
ASSUMPTION Assume the statement is true for n = k
𝑓 (𝑘 )=𝑘3−7𝑘+9𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒𝑏𝑦 3 𝑓𝑜𝑟 𝑘∈ℤ+¿ ¿
INDUCTIVE Show that the statement is then true for n = (k + 1)𝑓 (𝑘+1 )=(𝑘+1)3−7(𝑘+1)+9
𝑓 (𝑘+1 )=𝑘3+3𝑘2+3𝑘+1−7𝑘−7+9
𝑓 (𝑘+1 )=𝑘3+3𝑘2−4𝑘+3
Multiply out the brackets
Group up terms
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible
by 3 for all positive integers
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
INDUCTIVE Show that the statement is then true for n = (k + 1)
𝑓 (𝑘+1 )=𝑘3+3𝑘2−4𝑘+3𝑓 (𝑘 )=𝑘3−7𝑘+9Subtract f(k) from f(k + 1), using the expressions above
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿(𝑘3+3𝑘2−4𝑘+3 )− (𝑘3−7𝑘+9 )
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿𝑘3+3𝑘2−4𝑘+3−𝑘3+7 𝑘−9
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿3𝑘2+3𝑘−6
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿3 (𝑘2+𝑘−2)
𝑓 (𝑘+1 )=¿𝑓 (𝑘 )+3(𝑘2+𝑘−2)
‘Remove’ the brackets
Group
termsTake out 3 as a
factor
Add f(k)
This shows that f(k + 1) is just f(k) with an expression added on
We assumed f(k) was divisible by 3 The expression to be added is divisible by 3
So the answer must be divisible by 3, if f(k) is! As the first case (n = 1) was divisible by 3, the
statement must be true!
CONCLUSION
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible
by 133 for all positive integers
This example will require more manipulation as we work through it, but is essentially the same as the
previous two…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
BASIS Show that the statement is true for n = 1𝑓 (𝑛)=11𝑛+1+122𝑛−1
Sub in n = 1𝑓 (1 )=112+121
¿133Calculate
133 is divisible by 133, so the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible
by 133 for all positive integers
This example will require more manipulation as we work through it, but is essentially the same as the
previous two…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
ASSUMPTION Assume the statement is true for n = k
𝑓 (𝑘 )=11𝑘+1+122𝑘−1𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒𝑏𝑦 133 𝑓𝑜𝑟 𝑘∈ℤ+¿ ¿
INDUCTIVE Show that the statement is then true for n = (k + 1)
𝑓 (𝑘+1 )=11(𝑘+1 )+1+122 (𝑘+1)− 1
𝑓 (𝑘+1 )=11𝑘+2+122𝑘+1
𝑓 (𝑘+1 )=11×11𝑘+1+122×122𝑘−1
𝑓 (𝑘+1 )=11 (11𝑘+1)+144 (122𝑘−1)
Simplify powers
11k+2 = 11 x 11k+1122k+1 = 122 x 122k-1
Simplify
Rewrite
*
*They have been re-written in this way so that, on
the next step, the 11s and 12s have the same powers as in the f(k) expression and therefore can
be grouped up!
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible
by 133 for all positive integers
This example will require more manipulation as we work through it, but is essentially the same as the
previous two…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
INDUCTIVE Show that the statement is then true for n = (k + 1)
𝑓 (𝑘+1 )=11(11𝑘+1)+144 (122𝑘−1)𝑓 (𝑘 )=11𝑘+1+122𝑘−1
Subtract f(k) from f(k + 1), using the expressions above
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿[11 (11𝑘+1)+144 (122𝑘−1) ]− [11𝑘+1+122𝑘−1 ]
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿10(11𝑘+1)+143(122𝑘− 1)
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=¿10 (11𝑘+1 )+10 (122𝑘− 1 )+133 (122𝑘−1 )
𝑓 (𝑘+1 )− 𝑓 (𝑘 )=10 𝑓 (𝑘)+133 (122𝑘−1)
𝑓 (𝑘+1 )=11 𝑓 (𝑘 )+133(122𝑘− 1)
Group terms
Split the 143 into 2
partsThe first 2 terms are
just 10 lots of f(k)
Add f(k)
If f(k) is divisible by 133, so is 11f(k) 133(122k-1) is divisible by 133 Therefore f(k+1) will also be divisible by 133
As f(1) was divisible by 133, the statement is therefore true!
Make sure you practise enough so you can spot how and when to manipulate in this way!
Teachings for Exercise 6C
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
You will have seen recurrence relations in C1. A recurrence relation
is a sequence where generating a term relies on a previous term.
It is very important that you understand the notation!
6C
𝑈𝑛+1=𝑈𝑛+5Example 1
𝑈 1=3
The next term in the sequence
The current term
The first term
This is telling you that the first number in the sequence is 3
And to get the next number, you add on 5 The sequence will be: 3, 8, 13, 18, 23…… and so on… As this is an arithmetic sequence, we can use the
formula from C1 for the ‘nth’ term.. (a + (n -1)d)
𝑈𝑛+1=3𝑈𝑛−1Example 2
𝑈 1=1
This is telling you that the first number in the sequence is 1
To get the next number, you multiply the current number by 3 and subtract 1…
The sequence will be: 1, 2, 5, 14, 41, 122 ……and so on… This is NOT an arithmetic sequence, so we cannot use the
arithmetic sequence method for the nth term The ‘nth’ term for a sequence like this is far more
complicated! They can also be proven to be correct (once you think you
know what they are!) by use of induction!
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2.
Before we start, let’s generate the first 5 terms in both the
ways shown above…
You do not need to do this on an exam, this is just to show you
the two ways of generating the sequence give the same result!
So now, let’s prove this is the case!
6C
𝑢𝑛+1=3𝑢𝑛+4
𝑢1=1𝑢2=3 (1 )+4¿7
𝑢3=3 (7 )+4¿25
𝑢4=3 (25 )+4¿79
𝑢4=3 (79 )+4¿241
Using the recurrence relation
Using the ‘nth’ term formula𝑢𝑛=3
𝑛−2
𝑢1=31−2¿1
𝑢2=32−2¿7
𝑢3=33−2¿25
𝑢4=34−2¿79
𝑢5=35−2¿241
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2.
So we are being asked to show that, for the sequence with this
recurrence relation, that the nth term formula is 3n – 2…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
BASIS Show that the statement is true for n = 1 and n = 2𝑢𝑛+1=3𝑢𝑛+4 𝑢1=1 There is a slight difference here. As we are given u1 already (n = 1), we need to also check the statement is true for n = 2…𝑢𝑛+1=3𝑢𝑛+4
𝑢1=1
𝑢2=3𝑢1+4𝑢2=7
𝑢𝑛=3𝑛−2
𝑢1=31−2𝑢1=1
𝑢2=32−2𝑢2=7
The first 2 terms are both 1 and 7, so the statement is true for n = 1 and 2
We already know u1
Now use the recurrence relation to
find u2
Calculate
Sub in n = 1
Calculate
Now sub in n = 2
Calculate
𝑢𝑛=3𝑛−2
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2.
So we are being asked to show that, for the sequence with this
recurrence relation, that the nth term formula is 3n – 2…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
ASSUMPTION Assume that the statement is true for n = k𝐹𝑜𝑟 𝑛=𝑘 ,𝑢𝑘=3
𝑘−2𝑓 𝑜𝑟 𝑘∈ℤ+¿¿
𝑢𝑛+1=3𝑢𝑛+4
INDUCTIVE Use the recurrence relation to create an expression for uk+1𝑢𝑘+1=3𝑢𝑘+4
𝑢𝑘+1=3 (3𝑘−2)+4
𝑢𝑘+1=3𝑘+1−6+4
𝑢𝑘+1=3𝑘+1−2
Replace uk with the assumed expression
aboveMultiply out the brackets
Simplify
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2.
So we are being asked to show that, for the sequence with this
recurrence relation, that the nth term formula is 3n – 2…
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
𝑢𝑘=3𝑘−2
INDUCTIVE Use the recurrence relation to create an expression for uk+1
𝑢𝑘+1=3𝑘+1−2
The ‘k’ terms have all become ‘k + 1’
terms
CONCLUSION If the statement is true for ‘k’, it is also true for ‘k + 1’
We showed in the basis that it is true for n = 1 and n = 2
Therefore the statement is true for all
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
𝑢𝑛+2=5𝑢𝑛+1−6𝑢𝑛
The next term in the sequence
The current term
The previous
term
𝑢1=13𝑢2=35
The first term
The second term
So for this sequence, the next term is based on the current term AND the term before that!
This is why you have been given the first 2 terms…
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
BASIS Show that the statement is true for n = 1, n = 2 and n = 3
𝑢𝑛+2=5𝑢𝑛+1−6𝑢𝑛
𝑢1=13
The first 3 terms are 13, 35 and 97 for both sequences, so the statement has been shown to be true up to n = 3
We already know u1 and u2𝑢2=35
𝑢3=5(35)−6(13)
𝑢3=5𝑢2−6𝑢1 Sub in u2 and u1 to
find u3
𝑢3=97Calculate
𝑢𝑛=2𝑛+1+3𝑛+1
𝑢1=22+32
𝑢1=13
𝑢2=23+33
𝑢2=35
𝑢3=24+34
𝑢3=97
Calculate
Calculate
Calculate
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
ASSUMPTION Assume that the statement is true for n = k AND n = k + 1𝐹𝑜𝑟 𝑛=𝑘 ,𝑢𝑘=2
𝑘+1+3𝑘+1𝑓 𝑜𝑟 𝑘∈ℤ+ ¿¿
𝑢𝑛+2=5𝑢𝑛+1−6𝑢𝑛
𝐹𝑜𝑟 𝑛=𝑘+1 ,𝑢𝑘+1=2𝑘+2+3𝑘+2𝑓 𝑜𝑟 𝑘∈ℤ+ ¿¿
INDUCTIVE Use the recurrence relation to create an expression for uk+2𝑢𝑘+2=5𝑢𝑘+1−6𝑢𝑘
𝑢𝑘+2=5 (2𝑘+2+3𝑘+2)−6 (2¿¿𝑘+1+3𝑘+1)¿
𝑢𝑘+2=5 (2𝑘+2)+5 (3𝑘+2)−6 (2𝑘+1 )−6 (3𝑘+1 )
𝑢𝑘+2=5 (2𝑘+2)+5 (3𝑘+2)−3 (2𝑘+ 2 )−2 (3𝑘+2 )
6 (2𝑘+1 )¿3×2 (2𝑘+1 )¿3 (2𝑘+2 )
6 (3𝑘+1 )¿2×3 (3𝑘+1 )¿2 (3𝑘+2 )
Sub in the assumed expressions for uk+1 and uk
from beforeSplit the bracketed
parts upRewrite all as
powers of ‘k + 2’(see below)
6 = 3 x 2
The 2 adds 1 to the power
6 = 2 x 3
The 3 adds 1 to the power
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
ASSUMPTION Assume that the statement is true for n = k AND n = k + 1𝐹𝑜𝑟 𝑛=𝑘 ,𝑢𝑘=2
𝑘+1+3𝑘+1𝑓 𝑜𝑟 𝑘∈ℤ+ ¿¿
𝑢𝑛+2=5𝑢𝑛+1−6𝑢𝑛
𝐹𝑜𝑟 𝑛=𝑘+1 ,𝑢𝑘+1=2𝑘+2+3𝑘+2𝑓 𝑜𝑟 𝑘∈ℤ+ ¿¿
INDUCTIVE Use the recurrence relation to create an expression for uk+2𝑢𝑘+2=5𝑢𝑘+1−6𝑢𝑘
𝑢𝑘+2=5 (2𝑘+2+3𝑘+2)−6 (2¿¿𝑘+1+3𝑘+1)¿
𝑢𝑘+2=5 (2𝑘+2)+5 (3𝑘+2)−6 (2𝑘+1 )−6 (3𝑘+1 )
𝑢𝑘+2=5 (2𝑘+2)+5 (3𝑘+2)−3 (2𝑘+ 2 )−2 (3𝑘+2 )
Sub in the assumed expressions for uk+1 and uk
from beforeSplit the bracketed
parts upRewrite all as
powers of ‘k + 2’
𝑢𝑘+2=2 (2𝑘+2)+3 (3𝑘+2)
𝑢𝑘+2=2𝑘+3+3𝑘+ 3
Group the ‘like’ terms
The 2 and 3 add 1 to the powers of 2 and 3 respectively
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
CONCLUSION
𝑢𝑘=2𝑘+1+3𝑘+1
𝑢𝑘+1=2𝑘+2+3𝑘+2
𝑢𝑘+2=2𝑘+3+3𝑘+3
𝑢𝑘+1=2(𝑘+1 )+1+3 (𝑘+1) +1
𝑢𝑘+2=2(𝑘+2) +1+3(𝑘+2 )+1
These can both be written differently
As you can see, k is replaced with (k + 1), and then with (k + 2)
So we have shown that IF the statement is true for n = k and n = k + 1, then it must also be true for n = k + 2
As we showed in the basis that the statement is true for n = 1 and n = 2, then it must therefore be true for n = 3
And consequently it is then true for all values of n!
Teachings for Exercise 6D
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6D
[1 −10 2 ]
𝑛
=[1 1−2𝑛
0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
BASIS Show that the statement is true for n = 1
[1 −10 3 ]
𝑛
=[1 1−2𝑛
0 2𝑛 ]
[1 −10 2 ]
1
¿ [1 −10 2 ]
[1 1−21
0 21 ]¿ [1 −10 2 ]
Replace n with 1 Replace n with 1
Calculate
Calculate
So the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6D
[1 −10 2 ]
𝑛
=[1 1−2𝑛
0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
ASSUMPTION Assume the statement is true for n = k
[1 −10 2 ]
𝑘
=[1 1−2𝑘
0 2𝑘 ]𝑖𝑠𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑘∈ℤ+ ¿¿
INDUCTIVE Show using the assumption, that the statement will also be true for n = k + 1
[1 −10 2 ]
𝑘+1
¿ [1 −10 2 ]
𝑘
[1 −10 2 ]
1
¿ [1 1−2𝑘
0 2𝑘 ][1 −10 2 ]
Replace the power ‘k’ term with the assumed
matrix The second matrix doesn’t need the power!
Now we need to multiply these matrices using the skills from chapter 4!
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6D
[1 −10 2 ]
𝑛
=[1 1−2𝑛
0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
INDUCTIVE Show using the assumption, that the statement will also be true for n = k + 1
[1 −10 2 ]
𝑘+1
¿ [1 −10 2 ]
𝑘
[1 −10 2 ]
1
¿ [1 1−2𝑘
0 2𝑘 ][1 −10 2 ]
Replace the power ‘k’ term with the assumed
matrix The second matrix doesn’t need the power!
(1×1 )+((1−2𝑘)×0) (1×−1 )+((1−2𝑘)×2)(0×1 )+(2𝑘×0) (0×−1 )+(2𝑘×2)
1 −1+2−2(2𝑘)0 2(2𝑘)
¿ [1 1−2𝑘+1
0 2𝑘+1 ]
1 1−2𝑘+1
0 2𝑘+1
Work out each term
Simplify (remember to
manipulate the powers)
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[1 −10 2 ]
𝑛
=[1 1−2𝑛
0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
CONCLUSION
[1 −10 2 ]
𝑘
=[1 1−2𝑘
0 2𝑘 ]𝑖𝑠𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑘∈ℤ+¿¿
We assumed that:
Using this, we showed that:
[1 −10 2 ]
𝑘+1
=[1 1−2𝑘+1
0 2𝑘+1 ]As you can see, all the ‘k’ terms have been replaced with
‘k + 1’ terms
Therefore, IF the statement is true for one term, it will also be true for the next term, and so on…
As we already showed that the statement is true for n = 1, it is therefore true for all values of n!
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[−2 9−1 4 ]
𝑛=[−3𝑛+1 9𝑛
−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
BASIS Show that the statement is true for n = 1
[−2 9−1 4 ]
𝑛=[−3𝑛+1 9𝑛
−𝑛 3𝑛+1]
[−2 9−1 4 ]
1
¿ [−2 9−1 4 ]
[−3 (1 )+1 9(1)−(1) 3 (1 )+1]
Replace n with 1 Replace n with 1
Calculate
Calculate
So the statement is true for n = 1
¿ [−2 9−1 4 ]
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[−2 9−1 4 ]
𝑛=[−3𝑛+1 9𝑛
−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
ASSUMPTION Assume the statement is true for n = k
[−2 9−1 4 ]
𝑘=[−3𝑘+1 9𝑘
−𝑘 3𝑘+1]𝑖𝑠𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑘∈ℤ+¿¿
INDUCTIVE Show using the assumption, that the statement will then be true for n = k + 1
[−2 9−1 4 ]
𝑘+1
¿ [−2 9−1 4 ]
𝑘
[−2 9−1 4]
1
¿ [−3𝑘+1 9𝑘−𝑘 3𝑘+1 ][−2 9
−1 4 ]
Replace the power ‘k’ term with the assumed matrix
The second matrix doesn’t need the
power!
Now we need to multiply these matrices using the skills from chapter 4!
Simplify terms
(probably a good idea to
do in stages…)
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[−2 9−1 4 ]
𝑛=[−3𝑛+1 9𝑛
−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
6D
INDUCTIVE Show using the assumption, that the statement will also be true for n = k + 1
((−3𝑘+1 )×−2)+(9𝑘×−1)
[−2 9−1 4 ]
𝑘+1
¿ [−2 9−1 4 ]
𝑘
[−2 9−1 4]
1
¿ [−3𝑘+1 9𝑘−𝑘 3𝑘+1 ][−2 9
−1 4 ]
Replace the power ‘k’ term with the assumed matrix
The second matrix doesn’t need the
power!
((−3𝑘+1 )×9)+(9𝑘×4)(−𝑘×−2)+((3𝑘+1)×−1) (−𝑘×9)+((3𝑘+1)×4)
(6 𝑘−2)+(−9𝑘)(2𝑘)+(−3𝑘−1)
(−27𝑘+9)+(36𝑘)(−9𝑘)+(12𝑘+4)
−3 𝑘−2−𝑘−1
9𝑘+93𝑘+4
¿ [−3𝑘−2 9𝑘+9−𝑘−1 3 𝑘+4 ]
Simplify fully
This is the answer to the multiplication
[−2 9−1 4 ]
𝑘=[−3𝑘+1 9𝑘
−𝑘 3𝑘+1]
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[−2 9−1 4 ]
𝑛=[−3𝑛+1 9𝑛
−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿
6D
CONCLUSION We assumed that:
Using this, we showed that:
𝑖𝑠𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑘∈ℤ+¿¿
[−2 9−1 4 ]
𝑘+1
=[−3𝑘−2 9𝑘+9−𝑘−1 3𝑘+4 ]
¿ [−3 (𝑘+1 )+1 9(𝑘+1)−(𝑘+1) 3 (𝑘+1 )+1]
Each part of the matrix can be written differently…
(you will see why in a moment!)
If you compare this to the original matrix, you can see that all the ‘k’ terms have been replaced with
‘k + 1’ terms
So we have shown that if the statement is true for n = k, it will also be true for n = k + 1
As it was true for n = 1, it is also true for all positive values of k!
Summary• We have seen how to use proof by induction
• We have seen how to use it in situations regarding the summation of a series, tests of divisibility, recurrence relationships and matrices
• The four steps will always be the same, you will need to practice the ‘clever manipulation’ behind some of the inductive steps though!