Need help with a proof by induction, please

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The discussion focuses on proving by induction that the expression (3^(2n-1)) + 1 is divisible by 4 for all integers n. The base case for n=1 is established as 4, which is divisible by 4. The inductive step involves assuming the statement holds for n=k and then proving it for n=k+1 by rewriting the expression as 3^(2k-1) * 3^2 + 1. The proof hinges on demonstrating that this form maintains divisibility by 4, leveraging the assumption that a_k is divisible by 4.

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yelena0000
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Using proof by induction, prove that (3^(2n-1))+1 is divisible by 4

so this is what i could do so far:

for n=1
3^(2*1-1)+1=4 which is divisible by 4
assume truth for n=k
(3^(2k-1))+1 is divisible by 4
and i know that next i have to prove for n=k+1 but i really have no idea what to do witht that.
please help
 
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yelena0000 said:
and i know that next i have to prove for n=k+1 but i really have no idea what to do with that.

If you substitute n = k + 1, you get

[tex] 3^{2(k+1) - 1} + 1 = 3^{2k-1} 3^2 + 1[/tex]

By assumption, [tex]4 | 3^{2k-1} + 1[/tex] so try to re-write [tex]3^{2k-1} 3^2 + 1[/tex] in a form that has a factor of [tex]3^{2k-1} + 1[/tex].
 
The general term is

[tex] a_n = 3^{2n-1}+1[/tex]

and you've shown that $a_1$ is divisible by four, and you've assumed the same for $a_k$ for some $k \ge 1$.

Look at [tex]a_{k+1}[/tex].

[tex] 3^{2(k+1)-1} +1 = 3^{2k+2-1} + 1 = 3^{2k -1} 3^2 + 1[/tex]

The goal is to show that this is also divisible by four - the fact that [tex]a_k[/tex] is divisible by four will play a role in this.
 
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