Question

Use mathematical induction to prove that ((n^2)+5)n is divisible by 6 for all integers n>=1.

Answer 1

So, write it out carefully.

Let P(n) be the statement: ((n^2)+5)n is divisible by 6

You want to show that
Step 1. P(1) is true
Step 2. That P(k) ==> P(k+1) for all integers k greater than or equal to 1.
Step 3. Then conclude by the Principle of Mathematical Induction that P(n) is true for all n greater than or equal to 1.

Answer 2

Given those three steps, which one or ones are you having trouble with?

Answer 3

hmm

Answer 4

so for step one I have:

Answer 5

\[p(1): [((1)^{2}+5)(1)] / 6 MOD 0 ->6/6 MOD 0\]

Answer 6

Is it ok to use % (MOD) to prove this?

Answer 7

When I use mod = 0, I just think: remainder of 0

Answer 8

You here hero?

Answer 9

no, don't use mod. You're making it too complicated.

Answer 10

Just look at it. When n = 1, what does
((n^2)+5)n
equal?

Answer 11

6

Answer 12

so it is divisible for p(1)

Answer 13

Right and obviously 6 is divisible by 6 so the statement P(1) is true.

Answer 14

so assume p(n) is divisivle by 6

Answer 15

Then prove p(n)=>p(n+1)

Answer 16

Step 2: Now suppose P(k) for k >= 1. I.e.,
k(k^2 + 5) = 6m for some integer m

Answer 17

Now you want to show that
(k+1)( (k+1)^2 + 5 ) = 6p
for some other integer p using what I just wrote down.

Answer 18

This shows that P(k) ==> P(k+1)

Answer 19

Right

Answer 20

So now manipulate (k+1)( (k+1)^2 + 5 )

Answer 21

So what i've done so far is facotr and foil and what not

Answer 22

to get:

Answer 23

\[n ^{3}+2n ^{2}+6n+6\]

Answer 24

No

Answer 25

and write it in k.

Answer 26

ok

Answer 27

Did I make an error, or should I not multiply out

Answer 28

Yes you made an error. But in any case, you want to exploit terms that look like
k(k^2 + 5)
so completely expanding out might not be the best strategy.

Answer 29

So if k = 6m , where m is an integer, then k is divisible by 6

Answer 30

No, that's not what P(k) says. P(k) says that
k(k^2 + 5) is divisible by 6.

That is
k(k^2 + 5) = 6m for some integer m

Answer 31

Oh i didnt mean our k

Answer 32

I just mean in general, if something = 6(something else)

Answer 33

Now you want to show that P(k+1) is true. That is, you want to show that
(k+1)( (k+1)^2 + 5 ) is divisible by 6
i.e.,
(k+1)( (k+1)^2 + 5 ) = 6p for some integer p.

Answer 34

then something is divisible by 6?

Answer 35

yes

Answer 36

If a and b are integers and a = 6b, then a is divisible by 6.

Answer 37

hm, ok

Answer 38

I'll leave you in Hero's capable hands or someone else. I've got some other things to do.

Answer 39

Is hero here?

Answer 40

hmm

Answer 41

we've shown p(1) is true

Answer 42

Assume \[p(k): (k^{2}+5)k=6m\] is true

Answer 43

show p(k) => p(k+1)

Answer 44

\[p(k+1): ((k+1)^{2}+5)(k+1) = 6m\]

Answer 45

Any tip Zarkon? Im stuck

Answer 46

multiply out the (k+1)^2

Answer 47

and stop writing that it is equal to 6m...since that is what we want to show

Answer 48

right

Answer 49

\[p(k+1): (k ^{2}+2k+6)(k+1)\]

Answer 50

write is as \[(k ^{2}+5+2k+1)(k+1)\]

Answer 51

\[[(k ^{2}+5)+(2k+1)](k+1)\]

Answer 52

now multiply this out ...but keep the \[k^2+5\] as together

Answer 53

\[(k ^{2}+5)k+(k ^{2}+5)+(2k+1)k+(2k+1)\]

Answer 54

expand this part ... \[(k ^{2}+5)+(2k+1)k+(2k+1)\]

Answer 55

\[3k ^{2}+3k+6\]

Answer 56

yes

Answer 57

so you have
\[(k ^{2}+5)k+3k ^{2}+3k+6\]

write as

\[(k ^{2}+5)k+3k(k+1)+6\]

Answer 58

what can you tell me about the above statement

Answer 59

cubic...

Answer 60

not enough to say it = 6m, which is what we want right?

Answer 61

what doe we know about \[(k ^{2}+5)k\]

Answer 62

=6m

Answer 63

what about

6

Answer 64

ah yea

Answer 65

what about \[3k(k+1)\]

Answer 66

so we need to show 3k(k+1) = 6m

Answer 67

which is easy

Answer 68

3(1)((1)+1) = 3(2) = 6

Answer 69

\[k(k+1)\] is always even

Answer 70

So that is sufficient proof

Answer 71

Ok

Answer 72

thus \[2|k(k+1)\Rightarrow 6|3k(k+1)\]

Answer 73

2 Divides? whats that for?

Answer 74

we only need to show 6 divides

Answer 75

I'm just pointing out why we get that 6 divides what we want

Answer 76

So it would be

Answer 77

Thus \[(k ^{2}+5)k = 6m \rightarrow ((k+1)^{2}+5)(k+1) = 6m\]

Answer 78

Where the arrow means implies

Answer 79

I wouldn't use the same letter m

Answer 80

Im not going to use that at all im just going to say is divisible by 6

Answer 81

Is that ok to do?

Answer 82

sure

Answer 83

Thanks for the help

Answer 84

np

Answer 85

Mind checking out a solved question : http://openstudy.com/#/updates/4ee24839e4b0a50f5c5629c8