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?
When we say that 8 divides a mathematical expression, it means that the expression can be evenly divided by 8 without any remainder. In other words, the expression is a multiple of 8.
To prove that 8 divides this expression for all n>0, we can use mathematical induction. First, we can show that the expression is divisible by 8 when n=1. Then, we can assume that the expression is divisible by 8 for some arbitrary value of n=k. Using this assumption, we can show that the expression is also divisible by 8 when n=k+1. This completes the proof by induction, showing that the expression is divisible by 8 for all n>0.
Yes, there are other methods that can be used to prove this statement. For example, we could use the division algorithm to show that the remainder when dividing the expression by 8 is always 0. We could also use modular arithmetic to show that the expression is congruent to 0 mod 8 for all values of n>0.
Proving that 8 divides this expression for all n>0 is important because it helps us to better understand the properties of numbers and their relationships. It also allows us to make more accurate calculations and predictions in various mathematical and scientific fields.
Yes, there are many real-world applications of this proof. For example, in computer science and coding, this proof can be used to optimize algorithms and improve efficiency in calculations involving large numbers. It can also be applied in physics and engineering to accurately model and predict systems that involve exponential growth or decay.