# Prove by Induction the 11^(n+1)+12^(2n-1) is dvisible by 133.

• teroenza
In summary: This was helpful as I had not been able to do it before. I multiplied out and got the n+1 case as a multiple of 133.In summary, the problem involves proving that the expression 11^(n+1)+12^(2n-1) is divisible by 133 using induction. The base case of n=1 has been shown to hold, and the assumption that the original statement holds has been made. The next step is to show that the statement holds for n=n+1 by factoring out 11*12^2 and adding or subtracting the necessary multiples of 11^(n+2) and 12^(2n+1) to recreate the (n+1) form. By substit
teroenza

## Homework Statement

Prove by Induction the 11^(n+1)+12^(2n-1) is visible by 133.

For all n>1 or n=1

## The Attempt at a Solution

I have shown that the base case of n=1 holds.
Then I assumed that the original statement "11^(n+1)+12^(2n-1)is divisible by 133" holds.

Now I need to show the statement holds for n=n+1 ---> 11^(n+2)+12^(2n+1) must be shown to be divisible by 133. I see it equals

11^(n+1)*11+12^(2n-1)*12^2

I pull out 11*12^2 to create something times the original statement (which is divisible)

11*12^2*[11^(n+1)+12^(2n-1)]

multiplied out this equals 12^2*11^(n+2) + 11*12^(2n+1)

I then subtract from this the necessary multiples of 11^(n+2) and 12^(2n+1) to recreate(re-balance) the (n+1) form 6 lines above.

the two terms I subtract are -143*11^(n+2) - 10*12^(2n+1).

If I can show these are divisible by 133, I have completed the proof. I have checked via computer that their sum is divisible for n=1 through 15, but can't seem to find a way to show it.

teroenza said:

## Homework Statement

Prove by Induction the 11^(n+1)+12^(2n-1) is visible by 133.

For all n>1 or n=1

## The Attempt at a Solution

I have shown that the base case of n=1 holds.
Then I assumed that the original statement "11^(n+1)+12^(2n-1)is divisible by 133" holds.

Now I need to show the statement holds for n=n+1 ---> 11^(n+2)+12^(2n+1) must be shown to be divisible by 133. I see it equals

11^(n+1)*11+12^(2n-1)*12^2

I pull out 11*12^2 to create something times the original statement (which is divisible)

11*12^2*[11^(n+1)+12^(2n-1)]
You can't do that! Those expressions are not equivalent. If they were equivalent, you would be done here.
multiplied out this equals 12^2*11^(n+2) + 11*12^(2n+1)

I then subtract from this the necessary multiples of 11^(n+2) and 12^(2n+1) to recreate(re-balance) the (n+1) form 6 lines above.

the two terms I subtract are -143*11^(n+2) - 10*12^(2n+1).

If I can show these are divisible by 133, I have completed the proof. I have checked via computer that their sum is divisible for n=1 through 15, but can't seem to find a way to show it.

That is why I then add

-143*11^(n+2) - 10*12^(2n+1)

to recreate it [11^(n+1)*11+12^(2n-1)*12^2]

though now with the intent of showing that all the terms are divisible by 133. It is an equivalent statement when the negative terms are attached. Apologies if I did not make my method clear.

teroenza said:
That is why I then add

-143*11^(n+2) - 10*12^(2n+1)

to recreate it [11^(n+1)*11+12^(2n-1)*12^2]

though now with the intent of showing that all the terms are divisible by 133. It is an equivalent statement when the negative terms are attached. Apologies if I did not make my method clear.
... but then you cannot factor out 11*122 which is what the next line shows.

When I pulled the 11*12^2 out front, I did not legitimately factor it out. I just wrote it as that product times the original statement which I am assuming to be divisible. Because this was not a legitimate operation I then multiply out

11*12^2*[11^(n+1)+12^(2n-1)]

and see what I need to add or subtract back to make it equal the original (n+1) quantity

11^(n+1)*11+12^(2n-1)*12^2.

So I end up with this (the line above) is equal to

11*12^2*[11^(n+1)+12^(2n-1)] -143*11^(n+2) - 10*12^(2n+1)

where that in bold can be assumed by the original assumption to be divisible by 133, then if I can show the non-bold portion is also, I can say that the original (n+1) quantity is divisible by 133.

You're trying too hard, I think, and you aren't taking advantage of the fact that 11^(n+1)+12^(2n-1) is divisible by 133.

Hint: For 11^(n+1)+12^(2n-1) to be divisible by 133 means there exists some integer kn such that 11^(n+1)+12^(2n-1) = 133*kn. This in turn means that 11^(n+1) = 133*kn - 12^(2n-1).

Thank you both. I substituted for 12^(n-1)*12^2 and was able to write both terms as multiples of 133.

## 1. "What is induction?"

Induction is a mathematical proof technique used to prove that a statement holds for all natural numbers. It involves proving that the statement holds for a base case, typically n = 1, and then showing that if the statement holds for n, it also holds for n+1.

## 2. "How do you use induction to prove a statement?"

To prove a statement by induction, you must first prove that it holds for the base case. Then, you assume that the statement holds for some arbitrary value of n, and use this assumption to show that it also holds for n+1. This completes the inductive step and proves the statement for all natural numbers.

## 3. "What is the statement that we are trying to prove using induction?"

The statement we are trying to prove using induction is that 11^(n+1)+12^(2n-1) is divisible by 133 for all natural numbers n.

## 4. "Why is it necessary to prove a statement by induction?"

Proving a statement by induction is necessary because it guarantees that the statement holds for all natural numbers, not just a few specific cases. This is important in mathematics because it allows us to make generalizations and draw conclusions about a larger set of numbers.

## 5. "How can we apply induction to prove that 11^(n+1)+12^(2n-1) is divisible by 133?"

To apply induction, we would first prove that the statement holds for n = 1. Then, we would assume that the statement holds for some arbitrary value of n and use this assumption to prove that it also holds for n+1. Specifically, we would show that if 11^(n+1)+12^(2n-1) is divisible by 133, then 11^(n+2)+12^(2n+1) is also divisible by 133. This completes the inductive step and proves the statement for all natural numbers.

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