Question

prove that for every n>=1, the number 11^n+1 + 12^2n-1 is divisible by 133. ( using math induction)

Answer 1

Prove for n=1 first, to see whether it is true.

Answer 2

Assume the formula is true for n=k. Write that expression in terms of k. Then prove for n=k+1.

Answer 3

Which step are you stuck at the moment?

Answer 4

i know how to do the first two step. its just the last part

Answer 5

it just takes factoring

Answer 6

Note: \(133+11=144\)

Answer 7

hence \(144-11=12^2-11=133\)... do you see what to do now?

Answer 8

I tried doing thid:
133/ 11^k+2 + 12^2k+1 = 11*11^k+1 +12^2 * 12^2k-1

Answer 9

Presume \(11^{n+1}+12^{2n-1}=133k\) for some integers \(n,k\).

Answer 10

Now observe:$$11^{(n+1)+1}+12^{2(n+1)-1}=11\times11^{n+1}+12^2\times12^{2n-1}$$

Answer 11

$$11\times11^{n+1}+144\times12^{2n-1}=11\times11^{n+1}+(11+133)\times12^{2n-1}=\cdots$$

Answer 12

Using what oldrin.bataku wrote down above, you prove:
For n=k+1.
\[\large LHS=11^{k+2}+12^{2k+1}\]
\[\large =11(11^{k+1})+12^{2}(12^{2k+1})\]
Then you factor out the 11 and ignore the second term and copy down what you wrote for n=k.

Answer 13

By doing that, you have to RIG the expression.

Answer 14

$$11\times11^{n+1}+11\times 12^{2n-1}+133\times12^{2n-1}=11\times(11^{n+1}+12^{2n-1})+133\times12^{2n-1}$$

Answer 15

so$$11\times133k+133\times12^{2n-1}=133(11k+12^{2n-1})$$ which is clearly divisible by \(13\)

Answer 16

133*

Answer 17

Couldn't have said it better myself @oldrin.bataku

Answer 18

so i first do the base cake n=1
133/ 11^2 + 12
133/ 133 true
assume n = k
133/ 11^k+1 +12^2k-1 tru

Answer 19

Why are you assuming n = k... just presume there exists some \(n\) such that our conjecture holds and then show \(n+1\) also holds.

Answer 20

No you should write an expression. Write a formula like oldrin did then.

Answer 21

shouldn't*

Answer 22

I thought that one has to first do the base case, then assume true for n=k and then prove for n= k+1

Answer 23

And it's better to assume that n=k oldrin. SO you don't get confused with n=n.

Answer 24

@mathtutoring22 That's a good way of doing mathematical induction. Continue doing that.

Answer 25

@mathtutoring22 the \(k\) I used in mine is indeterminate... it's just there to state that our integer is divisible by \(133\)

Answer 26

When you assume the formula is true for n=k, you should write this down:
\[11^{k+1} +12^{2k-1}=133m\] where m is an integer.

Answer 27

oh okay that's fine

Answer 28

You can use any old letter. Just state that it is an integer etc. when you do.

Answer 29

how should i put this all together?

Answer 30

Well you should put all that chronologically and then show us so we can instantly tell what you need to fix etc.

Answer 31

@mathtutoring22 Presume for some positive integer \(k\) we have $$11^{k+1}+12^{2k-1}=133m$$for some integer \(m.\) Observe:$$11^{(k+1)+1}+12^{2(k+1)-1}=11\times11^{k+1}+12^2\times12^{2k-1}=\dots$$Clearly, \(11^{1+1}+12^{2(1)-1}=11^2+12=121+12=133\) is divisible by \(133\). Hence \(11^{k+1}+12^{2k-1}\) is divisible by \(133\) for all \(k\ge1\).

as an example...

Answer 32

Base Case?

Answer 33

idk why im so lost

Answer 34

I don't like following a rigid proof template... just make sure the logic checks out

Answer 35

@mathtutoring22 Just write down everything for this proof and we will tell you what you did wrong and we can patch things up for you. WE can't do that unless you show us what you've done.

Answer 36

like
Base case:
....
Assume true for n= k
...
Prove true for n= k+1

Answer 37

And then write your conclusion. You've got everything in the right order. I don't see anything wrong with that. So why are you making a fuss of being confused?

Answer 38

this is what i did:
Base case:
133/ 11^2 +12
133/133 true

Assume true for n=k
133/ 11^(k+1) +12^2k-1

prove for n= K+1

133/ 11^k+2 + 12^2k+2 = 11*11^k+1 + 12^2 * 12^2k-1 = 11*11^11^k+1 + (133+11)*12^12^2k-1 = 11*11^k+1*133*^12^2k-1 + 11*12^2k-1 =
11*(11^K+1+12^2k-1) <-- Inductive h. + 133*12^2k-1

Answer 39

when you assume true for n=k. You should write a formula. It's much clearer than just writing an expression. For example:
\[11^{k+1} +12^{2k-1}=133m where m is an integer.

Answer 40

\[11^{k+1} +12^{2k-1}=133m\] where m is an integer.

Answer 41

So when it comes to proving for n=k+!, you can write down LHS [Left Hand SIde]=

Answer 42

and then go down the page until you reach the last line that you do which will be:
\[=133[11m+{12^2k-1}]\]
\[=133M\] where M [This M is different to the m before!!] is an integer.

Answer 43

n=k+1*

Answer 44

\[=133[11m+12^{2k-1}\]*

Answer 45

Sorry about the typo there with the second last line. The website is a bit slow when it comes to typing.

Answer 46

]**

Answer 47

can you show me the first two line of the last step?

Answer 48

Okay well you wrote what n=k was. You just substitute n for k. You would do the same thing for n=k+!. Your original equation was:
\[11^{n+1} +12^{2n-1}=133m\]
Then your very first step when proving n=k+! would be basically substituting n with k+1:
\[LHS=11^{(k+1)+1}+12^{2(k+1)-1}\]

Answer 49

n=k+1*

Answer 50

then your second line would be trying to take out the extra terms so that you can put it in the form of n=k: which was this-
\[LHS=11^{k+1} +12^{2k-1}\]
SO your SECOND step would be to follow your first step:
\[=(11\times 11^{k+1})+(12^2\times12^{2k-1})\]

Answer 51

You're trying to put it in the form of n=k. That's the first two lines. Now you would be "rigging" the equation until you get it in the same form of 133m. And the last two lines I showed you.

Answer 52

Thanks

Answer 53

No worries.