
#1
Mar2304, 05:10 PM

P: 106

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. 



#2
Mar2304, 05:20 PM

P: 1,006

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] 



#3
Mar2304, 07:36 PM

P: 106

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 



#4
Mar2404, 01:36 AM

P: 1,006

Proof by Induction
[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. 



#5
Mar2404, 02:30 AM

P: 406

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] 



#6
Mar2404, 05:44 AM

P: 1,006

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). 



#7
Oct1404, 01:30 PM

P: 1

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! 


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