The problem asks that I compute [tex]3^n-2^n[/tex] for positive integer values of n, starting at 1 and working through a handful of consecutive integers.

From there I am to make a general observation about the values and then attempt to prove it with induction.

So I found the values:

n=1, 3^1-2^1 = 1

n=2, 3^2-2^2 = 5

n=3, 3^3-2^3 = 19

n=4, 3^4-2^4 = 65

n=5, 3^5-2^5 = 211

n=6, 3^6-2^6 = 665

n=7, 3^7-2^7 = 2059

n=8, 3^8-2^8 = 6305

n=9, 3^9-2^9 = 19171

From this I made the observation that for even values of n, [tex]3^n-2^n[/tex] is divisible by 5.

So I worked with the idea that [tex]3^{2n}-2^{2n}[/tex] is divisible by 5 for ALL positive integers.

Using mathematical induction:

For n=1

[tex]3^{2(1)}-2^{2(1)} = 5[/tex]

For k

[tex]3^{2k}-2^{2k}[/tex]

For k+1

[tex]3^{2(k+1)}-2^{2(k+1)}[/tex]

[tex]= 3^{2(k+1)}-2^{2(k+1)} -3^{2k} +3^{2k} -2^{2k} +2^{2k}[/tex]

[tex]= 3^{2k}(3^2-1) - 2^{2k}(2^2-1) + (3^{2k} -2^{2k})[/tex]

The 3rd term is case k, assumed to be divisible by 5

Working with the remaining terms:

[tex]3^{2k}(8) - 2^{2k}(3)[/tex]

[tex]= 3^{2k}(8) - 2^{2k}(3) + 2^{2k}(5) - 2^{2k}(5)[/tex]

[tex]= 3^{2k}(8) - 2^{2k}(8) + 2^{2k}(5)[/tex]

[tex]= 8(3^{2k} - 2^{2k}) + 5(2^{2k})[/tex]

The first term has case k as a factor, and the second term has 5 as a factor, making both divisible by 5

Thus [tex]3^{2n}-2^{2n}[/tex] is divisible by 5 for all positive integers

Does this look valid?

Any way to clean it up?

Thanks in advance for your time!

GM

From there I am to make a general observation about the values and then attempt to prove it with induction.

So I found the values:

n=1, 3^1-2^1 = 1

n=2, 3^2-2^2 = 5

n=3, 3^3-2^3 = 19

n=4, 3^4-2^4 = 65

n=5, 3^5-2^5 = 211

n=6, 3^6-2^6 = 665

n=7, 3^7-2^7 = 2059

n=8, 3^8-2^8 = 6305

n=9, 3^9-2^9 = 19171

From this I made the observation that for even values of n, [tex]3^n-2^n[/tex] is divisible by 5.

So I worked with the idea that [tex]3^{2n}-2^{2n}[/tex] is divisible by 5 for ALL positive integers.

Using mathematical induction:

For n=1

[tex]3^{2(1)}-2^{2(1)} = 5[/tex]

For k

[tex]3^{2k}-2^{2k}[/tex]

For k+1

[tex]3^{2(k+1)}-2^{2(k+1)}[/tex]

[tex]= 3^{2(k+1)}-2^{2(k+1)} -3^{2k} +3^{2k} -2^{2k} +2^{2k}[/tex]

[tex]= 3^{2k}(3^2-1) - 2^{2k}(2^2-1) + (3^{2k} -2^{2k})[/tex]

The 3rd term is case k, assumed to be divisible by 5

Working with the remaining terms:

[tex]3^{2k}(8) - 2^{2k}(3)[/tex]

[tex]= 3^{2k}(8) - 2^{2k}(3) + 2^{2k}(5) - 2^{2k}(5)[/tex]

[tex]= 3^{2k}(8) - 2^{2k}(8) + 2^{2k}(5)[/tex]

[tex]= 8(3^{2k} - 2^{2k}) + 5(2^{2k})[/tex]

The first term has case k as a factor, and the second term has 5 as a factor, making both divisible by 5

Thus [tex]3^{2n}-2^{2n}[/tex] is divisible by 5 for all positive integers

Does this look valid?

Any way to clean it up?

Thanks in advance for your time!

GM

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