Prove that for all integers n > 0, 8n – 3n is divisible by 5. check my answer?

- anonymous

Prove that for all integers n > 0, 8n – 3n is divisible by 5. check my answer?

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- 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

I think you meant to say `Induction Hypothesis: 8^k-3^k is divisible by 5`

- 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

what is the significance of 5q??

- jim_thompson5910

it just means "5 times some integer"

- jim_thompson5910

if I said "100 is divisible by 5" I can state that as 100 = 5q
In this case, q = 20

- anonymous

ok, yeah. so that just means divisible by 5

- jim_thompson5910

yeah

- 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

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

i do not understand.8^(k+1) is just 64, no? not 64^K?

- jim_thompson5910

how are you getting 64? you can't replace k with 1

- anonymous

i didn't know i couldn't do that.

- 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

do you see how I turned `8^(k+1)-3^(k+1)` into `8*8^k-3*3^k` ?

- anonymous

no, that was my next question. i thought k+1 would turn into ^2

- jim_thompson5910

recall that
\[\LARGE x^{y+z} = x^y * x^z\]

- jim_thompson5910

so \[\Large 8^{k+1} = 8^k*8^1 = 8^k*8 = 8*8^k\]
make sense?

- anonymous

yes. thank you. and now i have to do the same thing for the other side, right?

- jim_thompson5910

for the 3^(k+1) part you mean? yes

- anonymous

ok, so i just get 3^k*3+5q=6k+5q

- 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

multiply

- anonymous

40q + 24^k - 9^k
40q + 15^k

- jim_thompson5910

8*3^k isn't equal to 24^k

- 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

im going to have to study that, tomorrow. can you look at a different kind of problem with me?

- jim_thompson5910

sure

- anonymous

1^3 + 2^3 + 3^3 + … + n^3 = n^2(n + 1)^2/4 for all positive integers n

- jim_thompson5910

I'm sure you have the base case already done?

- anonymous

no. i just started looking at it. i have not tried solving a sequence kind of problem yet.

- anonymous

n^2(n^2+1)/4

- jim_thompson5910

you'll use induction here

- anonymous

i have less idea about this problem then i did the last :|

- jim_thompson5910

the base case is n = 1

- jim_thompson5910

prove that is true

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