anonymous
  • anonymous
Prove that for all integers n > 0, 8n – 3n is divisible by 5. check my answer?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Base Case: n=1 8^1-3^1=5 5/5=1 Induction Hypothesis: 8^k-3^k is divisible by 3 8^(k+1)-3^(k+1) 8^(2)-3^(2) 64-9=53 53/3
jim_thompson5910
  • jim_thompson5910
I think you meant to say `Induction Hypothesis: 8^k-3^k is divisible by 5`
jim_thompson5910
  • jim_thompson5910
if we assume that claim is true, then we can say 8^k-3^k = 5q where q is some integer. Agreed?

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More answers

anonymous
  • anonymous
what is the significance of 5q??
jim_thompson5910
  • jim_thompson5910
it just means "5 times some integer"
jim_thompson5910
  • jim_thompson5910
if I said "100 is divisible by 5" I can state that as 100 = 5q In this case, q = 20
anonymous
  • anonymous
ok, yeah. so that just means divisible by 5
jim_thompson5910
  • jim_thompson5910
yeah
jim_thompson5910
  • jim_thompson5910
the goal is to prove that if 8^k-3^k = 5q is assumed to be true, then 8^(k+1)-3^(k+1) = 5s where q and s are integers
jim_thompson5910
  • jim_thompson5910
here are 2 hints: 1) isolate the `8^k` in `8^k-3^k = 5q` to get `8^k = 5q + 3^k` 2) break down `8^(k+1)-3^(k+1) ` into `8*8^k-3*3^k`
anonymous
  • anonymous
i do not understand.8^(k+1) is just 64, no? not 64^K?
jim_thompson5910
  • jim_thompson5910
how are you getting 64? you can't replace k with 1
anonymous
  • anonymous
i didn't know i couldn't do that.
jim_thompson5910
  • jim_thompson5910
you can replace k with 1 if you wanted to prove that specific piece. But we're trying to prove the whole thing in general (for any number k)
jim_thompson5910
  • jim_thompson5910
do you see how I turned `8^(k+1)-3^(k+1)` into `8*8^k-3*3^k` ?
anonymous
  • anonymous
no, that was my next question. i thought k+1 would turn into ^2
jim_thompson5910
  • jim_thompson5910
recall that \[\LARGE x^{y+z} = x^y * x^z\]
jim_thompson5910
  • jim_thompson5910
so \[\Large 8^{k+1} = 8^k*8^1 = 8^k*8 = 8*8^k\] make sense?
anonymous
  • anonymous
yes. thank you. and now i have to do the same thing for the other side, right?
jim_thompson5910
  • jim_thompson5910
for the 3^(k+1) part you mean? yes
anonymous
  • anonymous
ok, so i just get 3^k*3+5q=6k+5q
jim_thompson5910
  • jim_thompson5910
we have 8*8^k-3*3^k now replace the 8^k with 5q + 3^k 8*8^k-3*3^k 8*(5q + 3^k) - 3*3^k 8*5q + 8*3^k - 3*3^k do you see what's next?
anonymous
  • anonymous
multiply
anonymous
  • anonymous
40q + 24^k - 9^k 40q + 15^k
jim_thompson5910
  • jim_thompson5910
8*3^k isn't equal to 24^k
jim_thompson5910
  • jim_thompson5910
8*5q + 8*3^k - 3*3^k 5*8q + 8*3^k - 3*3^k 5*8q + (8*3^k - 3*3^k) 5*8q + (8 - 3)*3^k 5*8q + 5*3^k 5*(8q + 3^k) 5*s where s = 8q + 3^k and s is an integer. So this all proves that if 8^k - 3^k = 5q, then 8^(k+1)-3^(k+1) = 5s q and s are integers
anonymous
  • anonymous
im going to have to study that, tomorrow. can you look at a different kind of problem with me?
jim_thompson5910
  • jim_thompson5910
sure
anonymous
  • anonymous
1^3 + 2^3 + 3^3 + … + n^3 = n^2(n + 1)^2/4 for all positive integers n
jim_thompson5910
  • jim_thompson5910
I'm sure you have the base case already done?
anonymous
  • anonymous
no. i just started looking at it. i have not tried solving a sequence kind of problem yet.
anonymous
  • anonymous
n^2(n^2+1)/4
jim_thompson5910
  • jim_thompson5910
you'll use induction here
anonymous
  • anonymous
i have less idea about this problem then i did the last :|
jim_thompson5910
  • jim_thompson5910
the base case is n = 1
jim_thompson5910
  • jim_thompson5910
prove that is true

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