Question

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

Answer 1

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

Answer 2

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

Answer 3

if we assume that claim is true, then we can say

8^k-3^k = 5q

where q is some integer. Agreed?

Answer 4

what is the significance of 5q??

Answer 5

it just means "5 times some integer"

Answer 6

if I said "100 is divisible by 5" I can state that as 100 = 5q

In this case, q = 20

Answer 7

ok, yeah. so that just means divisible by 5

Answer 8

yeah

Answer 9

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

Answer 10

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`

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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)

Answer 15

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

Answer 16

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

Answer 17

recall that

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

Answer 18

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

make sense?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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?

Answer 23

multiply

Answer 24

40q + 24^k - 9^k

40q + 15^k

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

sure

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

you'll use induction here

Answer 34

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

Answer 35

the base case is n = 1

Answer 36

prove that is true