anonymous
  • anonymous
If n >= 1 is an integer, prove that 2^(2n) - 1 is divisible by 3. Is there a formal way to prove this or can I just demonstrate it by plugging in n = 1,2,3,etc?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Proving by "plugging in" would involve an infinite number of cases.
experimentX
  • experimentX
try using induction
anonymous
  • anonymous
My problem is the only induction I know is plugging examples in!

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anonymous
  • anonymous
OK, how did this happen: 4(2^(2n) - 1 + 1) - 1 = 4 - 1?
experimentX
  • experimentX
let it put it up again, \[ 2^{2n + 2} - 1 = 4 \cdot2^{2n} - 1 = 4 (2^{2n} - 1 + 1)- 1\] you know 3 divides 2^(2n) - 1 from assumption of induction. \[ 4 (2^{2n} - 1) + 4 - 1 = 4 \cdot (2^{2n }-1) + 3 \] since 3 divides both terms, it divides the whole thing.
experimentX
  • experimentX
also notice that \[ 2^{2n} = 4^n = (3 + 1)^n = 3 \times (\text{something}) + 1 \] you can use this to show that the reminder is 1
anonymous
  • anonymous
When I assume what I'm trying to prove to be true, what are the things I must do to make the whole argument valid?
experimentX
  • experimentX
do you know how to use induction?
anonymous
  • anonymous
this is my first problem in induction? Semester started this week.
anonymous
  • anonymous
avoiding induction, I will give another way, \[f(n)= 4^n-1 \] the values of the sum of the digits of f(n) takes a function represented by g(n): \[3n+1 \]thus 3n+1-1 is divisible by 3
experimentX
  • experimentX
Oh!! first put few values of n to show that it is true like, n=1,2,3 ... up to some number 'k' if it holds for k+1 assuming it holds for k, then it holds for any finite number.
experimentX
  • experimentX
@Ahmad1 seeing that made me remember \[ 4^n - 1 = (4 - 1)\cdot (1 + 4 + 4^2 + \dots + 4^{n-1})\] it holds trivially
anonymous
  • anonymous
I don't understand this "the values of the sum of the digits of f(n) takes a function represented by g(n):3n+1"
anonymous
  • anonymous
@Ahmad1 what did I miss?
experimentX
  • experimentX
that is just another way of looking at the problem 4^n -1 = 3k 4^n = 3k + 1 for some integer k
anonymous
  • anonymous
Oh I see. I get it now. Thanks everybody.
experimentX
  • experimentX
have seen this theorem? http://en.wikipedia.org/wiki/Fermat's_little_theorem
anonymous
  • anonymous
nope and I don't know modular arithmetic.

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