Proof by Induction

Caldus

How would I go about proving that 8^n - 3^n (n >= 1) is divisible by 5 using mathematical induction? I tried this but I do not think it is right:

First, prove that 8^1 - 3^1 is divisible by 5. 8^1 - 3^1 = 5, which is divisible by 5.

Second, prove that 8^(k+1) - 3^(k+1) is divisible by 5 if k = n. Notice that 8^(k+1) - 3^(k+1) =

8*(8^k) - 3^k(3). Based on the induction hypothesis, we already know that 8^k - 3^k is divisible by 5. So we end up with 24*(8^k - 3^k), which is always divisible by 5 because the term inside the parenthesis is already divisible by 5. Multiplying that by any number will not change the fact that it is divisible by 5.

Am I right here? Thanks.

Chen

How did you figure this:
[tex]8*8^k - 3*3^k = 24(8^k - 3^k)[/tex]
?? You were doing fine up until that point. What you need to do is this:
[tex]8*8^k - 3*3^k = 5*8^k + 3*8^k + 3*3^k = 5*8^k + 3(8^k - 3^k)[/tex]

Caldus

OK, I'm having trouble with this one as well. Can someone help me?

1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n(2n - 1)(2n + 1))/3

Chen

[tex]1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} = \frac{4n^3 - n}{3}[/tex]

So for [tex]k = n + 1[/tex]:
[tex]1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 + (2n + 1)^2 = \frac{(n + 1)(2n + 1)(2n + 3)}{3}[/tex]
[tex]\underline{1^2 + 3^2 + 5^2 + ... + (2n - 1)^2} + (2n + 1)^2 = \frac{4n^3 + 12n^2 + 11n + 3}{3}[/tex]
We already know it's true for [tex]n[/tex] so you can replace the underlined part:
[tex]\frac{4n^3 - n}{3} + (2n + 1)^2 = \frac{4n^3 - n}{3} + 4n^2 + 4n + 1 = \frac{4n^3 + 12n^2 + 11n + 3}{3}[/tex]
Multiply by 3:
[tex]4n^3 - n + 12n^2 + 12n + 3 = 4n^3 + 12n^2 + 11n + 3[/tex]
QED.

recon

I have seen a beautiful geometrical way of deriving the sum of cubes of numbers, i.e. 1^3 + 2^3 + ... + n^3 = (1+2+...n)^2.

I wonder if there are simple ways of deriving [tex]1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3}[/tex]

Chen

You can prove this:
[tex]1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3}[/tex]
With a bit of geometry, yes. If you draw squares with a side of 1, 3, 5, etc., one below the other, you can find the sum of their areas with a bit of manipulation, but it's not exactly simple (the idea is simple, the equations are a bit big though).

Bonk

I have a problem:
Knowing that [tex]1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n + 1)(2n + 1)}{6}[/tex]and that [tex]1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2 (n + 1)^2 } {4}[/tex], calculate [tex]1^4 + 2^4 + 3^4 + ... + n^4[/tex].
Please, help me!