# Eleven to the N

Pro Problems > Math > Logic > Proofs > Proof by Induction## Eleven to the N

Prove by induction that for every integer n ≥ 1, 11^{n} is one more than a multiple of ten.

*Note: Proof by induction is not the simplest method of proof for this problem, so an alternate solution is provided as well.*

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Problem by BogusBoy

## Solution

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Click here to assign this problem to your students.## Similar Problems

### Sum of Integers Proof

1 = 1 =

1 + 2 = 3 =

1 + 2 + 3 = 6 =

1 + 2 + 3 + 4 = 10 =

1 + 2 + 3 + 4 + 5 = 15 =
It appears from this that the sum of the first n positive integers is . Can you prove this by induction?

1(1 + 1)

2

1 + 2 = 3 =

2(2 + 1)

2

1 + 2 + 3 = 6 =

3(3 + 1)

2

1 + 2 + 3 + 4 = 10 =

4(4 + 1)

2

1 + 2 + 3 + 4 + 5 = 15 =

5(5 + 1)

2

n(n + 1)

2

### Multiples of Pi/4

Prove by induction that for any non-negative integer n,

tan ((4n + 1)p

4

*Additional Question: Could this induction be extended to all integers, not just negative ones? If so, how?*

*Note: Without rigorous proof, we can see that this is true, since the angles which match the equation are all in the first and third quadrant, and all have reference triangles of. Nevertheless, this is a good practice exercise in inductive proof.*

p

4

### Square Sum Proof

Prove by induction that the sum of the first n positive perfect squares is:

n(n + 1)(2n + 1)

6

### Series Proof

Use a proof by induction to prove that the first n terms of the series

1

2

1

4

1

8

1

2

^{n}2

^{n}- 12

^{n}### Product and Power

Prove by induction that for all integers n>3:

3^{n} > 9n

### Geometric Sum Proof

Give a proof by induction to show that for every non-negative integer n:

2^{0} + 2^{1} + 2^{2} + ... + 2^{n} = 2^{n + 1} - 1

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