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For all n E Z, n >= 0 prove that
(a.) 2^2n+1 + 1 is divisible by 3
(b.) n^3 +(n + 1)^3 + (n+2)^3 is divisible by 9
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These are both proofs by induction. First you prove the statement is true when n = 1, then you assume it's true for n = k and show that that means it's true for n = k+1. I'll do the first one.
a) Let p(n) be 2^(2n+1) + 1.
Then p(1) = 2^3 + 1 = 9, which is divisible by 3.
Now assume that p(k) is divisible by 3. So, 2^(2k+1) + 1 = 3t, for some integer t.
Consider p(k+1) = 2^(2(k+1)+1)) + 1.
This is 2^(2k+3) + 1
= 4 x 2^(2k+1) + 1
= (3+1) x 2^(2k+1) + 1
= 3 x 2^(2k+1) + 3t
= 3(2^(2k+1) + t)
Hence, p(k+1) is divisible by 3.
Then, by induction, p(n) is divisible by 3, for all n.
Why did the vector cross the road?
It wanted to be normal.
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you dont need induction for the second one, just expand the paranthesis and simplify, and you will end up with an expression that you can easy show that it is divisible by 9 by letting n=3 and n=3k±1. (you can also first change n^3 +(n + 1)^3 + (n+2)^3 to (n - 1)^3 + n^3 + (n+1)^3, which may simplify the expansion)
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This is 2^(2k+3) + 1
= 4 x 2^(2k+1) + 1
= (3+1) x 2^(2k+1) + 1
= 3 x 2^(2k+1) + 3t
= 3(2^(2k+1) + t)
I am trying to understand the proove. Where does the 4 comes from? How is it decomposed to get (3 + 1)?
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You add indices when mutliplying:
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