The discussion revolves around proving that 8 divides the expression 1 + 2 × 3^(n-1) + 5^n for all natural numbers n > 0. Participants explore the concept of mathematical induction as a method to establish this divisibility.
The discussion is ongoing, with participants actively working through the algebraic manipulation of the expression. Some guidance has been provided on factoring and identifying common terms, but there is still exploration of the reasoning behind the divisibility by 8.
Participants express confusion regarding the induction process and the steps taken in class, indicating a need for clarification on the induction hypothesis and the transition from n=k to n=k+1.
mtayab1994 said:I already checked for when n=1 and for that the induction is true because 8/8=1.
By the way is it written as the following?: S(n+1)-S(n)=1+2*3^(n-1)(3)+5^n(5)-S(n)
mtayab1994 said:I already checked for when n=1 and for that the induction is true because 8/8=1.
By the way is it written as the following?: S(n+1)-S(n)=1+2*3^(n-1)(3)+5^n(5)-S(n)
mtayab1994 said:After trying S(n+1)-S(n) i got the following:
-2+2*3^n+5^(n+1)-3^(n-1)-5^n
Sorry i typed it in wrong then i fixed it i gotDick said:First correct a couple of errors before you simplify. Where did the '-2' in front come from? Can you find the other error?
mtayab1994 said:Sorry i typed it in wrong then i fixed it i got
2*3^n+5^(n+1)-2*3^(n-1)-5^n
Dick said:Ok, now start trying to write that in a factored form. Combine the two powers of 3. Do the same for the two powers of 5.
mtayab1994 said:alright i factored out the powers with 5 and i got 5^n(5-1) but I'm having troubles factoring out the one's with 3's because one is multiplied by '2' and the other by '-2'
Dick said:Why would that be a problem? Just pull out the common factors of 2 and 3^(n-1)...
mtayab1994 said:what i got was 2(3^n)-2(3^(n-1))+5^n(5-1)
Dick said:Factor the 2(3^n)-2(3^(n-1)) part. I told you what to do in the last post. And there's nothing wrong with writing 4 instead of 5-1.
mtayab1994 said:ok i got 2(3^n-3^(n-1))+5^n(5-1)
Dick said:3^n=3*3^(n-1). There's still a common factor of 3^(n-1) you can pull out of the first term.
mtayab1994 said:can i write it as 2(3(3^(n-1)-3^(n-1)+5^n(4)
Dick said:Hmm. Write the first part as 2*3^(n-1)*(something). What's the something? That's what I mean by 'factor out the 3^(n-1)'.
mtayab1994 said:i'm sorry I'm getting no where because in class my teacher solves these sorts of problems by just multiplying each side or dividing each side of the equation so he can get some sort of K' which makes the induction true. This is one of the examples:
prove that 9 divides 4^n+6n-1
4(4^n+6n-1)=36k
4^(n+1)+24n-4=36k
4^(n+1)+18n+6n-9+5=36k
4^(n+1)+6n+5=36k-18k+9
4^(n+1)+6n+5=9(4k-2k+1)
4^(n+1)+6(n+1)-1=9k' and he got 9k' from 4k-2k+1
mtayab1994 said:And on top of that he never even showed us the induction step: n=k is true and we have to prove that n=k+1 is i had to find that all on my own.
Dick said:Ok, but we are getting off the track here. I'm trying to get you to show that S(n+1)-S(n)=(something)*3^(n-1)+4*5^n and what that (something) is. You got the 4 in front of the 5^n just fine. I'm not sure why are having problems with the 3^(n-1) part. And then go on from there to tell me why S(n+1)-S(n) is divisible by 8. Then try and tell me why that solves your induction problem.
mtayab1994 said:ok i got 2*3^(n-1)(3-1)+5^n(4)
mtayab1994 said:then i came out with 4(3^(n-1)+5^n)
mtayab1994 said:it's also divisible by 8 because 8 is a multiple of 4. So there for S(n-1)-S(n) is divisible by 8.
mtayab1994 said:yea i found the answer:
3^(n-1) is odd and so is 5^n and the sum of two odd numbers is an even number so therefore 8 divides any number 4k' for k' is an even number.
Dick said:Right. And you see how that makes the induction work, yes?