MHB Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7

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The discussion focuses on proving by induction that the expression 3^(2n+1) + 2^(n-1) is divisible by 7 for all integers n greater than or equal to 1. The base case has been established, and the inductive step involves showing that if the statement holds for n=k, it must also hold for n=k+1. The transformation of the expression during the inductive step is crucial, with participants noting the importance of rearranging the expression to highlight divisibility by 7. The discussion emphasizes the need for clarity in each step of the proof to ensure correctness. The problem-solving approach is centered on manipulating the expression to reveal its divisibility properties.
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Can someone with understanding of proof by induction help with this problem?

Prove by induction that 3 raised to 2n+1 + 2 raised to n-1 is divisible by 7 for all numbers greater than/or equal to 1. How do you do the inductive step?
 
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I have done the base case and some of the inductive..which I'm not sure I'm going in the right direction.
Inductive, So does it hold true for n=k+1
3 raised 2(k+1)+1 +2 raised(k+1)-1 = 3 raised 2k+2+1 +2 raised (k+1)-1
= 3 raised 2k+1 x 3 raised2 + 2 raised k x 2 raised 0
=9 x 3 raised 2k+1 + 1 x 2 raised k
= 27 x 3 raised 2k +1x2 raised k

Problem isn't posting correctly
 
$3^{2(k+1)+1}+2^{(k+1)-1}$​

$=\quad3^{2k+3}+2^k$

$=\quad9\cdot3^{2k+1}+2\cdot2^{k-1}$

$=\quad7\cdot3^{2k+1}+2\cdot\left(3^{2k+1}+2^{k-1}\right).$

It should be straightforward to proceed from here.

When doing problems of this kind, look at the number you want your expression to be divisible by (in this case $7$) and try and rearrange your expression to involve it.
 
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