proof by mathematical induction

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!

http://www.youtube.com/watch?v=tj7al6MXu7U



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






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







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







CONCLUSION

6A

∑𝑟=1

𝑛

(2𝑟−1 )=𝑛2


∑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!







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



Prove, by mathematical induction, that for ,

So we are now going to prove one of the formulae you have learnt in

chapter 5!


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


There will only be one term

here, that we get by subbing n = 1

into the expression

Calculate


Show that the statement is true for n = 1





chapter 5!


CONCLUSION

6A

∑𝑟=1

𝑛(𝑟 2)=1

6𝑛 (𝑛+1 ) (2𝑛+1 )


∑𝑟=1

𝑘(𝑟 2)=¿ ¿1+4+9+16………+𝑘2¿ 16 𝑘 (𝑘+1 ) (2𝑘+1 )



terms of k


will be equal to the expression above





chapter 5!


CONCLUSION

6A

∑𝑟=1

𝑛(𝑟 2)=1

6𝑛 (𝑛+1 ) (2𝑛+1 )


∑𝑟=1

𝑘(𝑟 2)=¿ ¿1+4+9+16………+𝑘2¿ 16 𝑘 (𝑘+1 ) (2𝑘+1 )


∑𝑟=1

𝑘+1

(𝑟 2)=¿ ¿1+4+9+16………+𝑘2+(𝑘+1)2


¿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!!





chapter 5!


CONCLUSION

6A

∑𝑟=1

𝑛(𝑟 2)=1

6𝑛 (𝑛+1 ) (2𝑛+1 )


∑𝑟=1

𝑘+1

(𝑟 2)=¿ ¿1+4+9+16………+𝑘2+(𝑘+1)2


¿16 𝑘 (𝑘+1 ) (2𝑘+1 )+(𝑘+1)2


¿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





chapter 5!


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!




This looks more complicated, but you just follow the same process as

you have seen already!


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


There will only be one term

here, that we get by subbing n = 1

into the expression

Calculate


Show that the statement is true for n = 1







CONCLUSION

6A

∑𝑟=1

𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]


∑𝑟=1

𝑛(𝑟 2𝑟 )=¿¿2+8+24+64……+𝑘(2𝑘)¿2 [1+(𝑘−1 )2𝑘 ]



terms of k


will be equal to the expression above







CONCLUSION

6A

∑𝑟=1

𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]


∑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


¿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)







CONCLUSION

6A

∑𝑟=1

𝑛(𝑟 2𝑟 )=2 [1+(𝑛−1)2𝑛 ]


¿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







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!


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






CONCLUSION

6B


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






CONCLUSION

6B

𝑓 (𝑘 )=32𝑘+11


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



Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible

by 3 for all positive integers


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







CONCLUSION

6B


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






CONCLUSION

6B


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



Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible


This example will require more manipulation as we work through it, but is essentially the same as the

previous two…


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







previous two…


CONCLUSION

6B


𝑓 (𝑘 )=11𝑘+1+122𝑘−1𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒𝑏𝑦 133 𝑓𝑜𝑟 𝑘∈ℤ+¿ ¿


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!






previous two…


CONCLUSION

6B


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!



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



relation


So we are being asked to show that, for the sequence with this

recurrence relation, that the nth term formula is 3n – 2…


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



relation





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



relation





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



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!


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…



relation





CONCLUSION

6C

BASIS Show that the statement is true for n = 1, n = 2 and n = 3


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



relation





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+ ¿¿


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



relation





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+ ¿¿


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



relation





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!


CONCLUSION

6D

[1 −10 2 ]

𝑛

=[1 1−2𝑛

0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿


[1 −10 3 ]

𝑛

=[1 1−2𝑛

0 2𝑛 ]

[1 −10 2 ]

1

¿ [1 −10 2 ]

[1 1−21

0 21 ]¿ [1 −10 2 ]


Calculate

Calculate

So the statement is true for n = 1





questions!


CONCLUSION

6D

[1 −10 2 ]

𝑛

=[1 1−2𝑛

0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿


[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!





questions!


CONCLUSION

6D

[1 −10 2 ]

𝑛

=[1 1−2𝑛

0 2𝑛 ]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿


[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)





questions!


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!




More complicated, but the same process!


CONCLUSION

[−2 9−1 4 ]

𝑛=[−3𝑛+1 9𝑛

−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿


[−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]


Calculate

Calculate

So the statement is true for n = 1

¿ [−2 9−1 4 ]






CONCLUSION

[−2 9−1 4 ]

𝑛=[−3𝑛+1 9𝑛

−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿


[−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…)






CONCLUSION

[−2 9−1 4 ]

𝑛=[−3𝑛+1 9𝑛

−𝑛 3𝑛+1]𝑓𝑜𝑟 𝑛∈ℤ+¿ ¿

6D


((−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]






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!

proof by mathematical induction

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