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

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The forum discussion centers on proving by induction that the expression \(3^{2n+1} + 2^{n-1}\) is divisible by 7 for all integers \(n \geq 1\). The user has successfully completed the base case and is working through the inductive step, specifically for \(n = k + 1\). The discussion highlights the importance of rearranging the expression to incorporate the divisor, 7, to facilitate the proof. The final expression derived is \(7 \cdot 3^{2k+1} + 2 \cdot (3^{2k+1} + 2^{k-1})\), which leads to the conclusion that the inductive step can be completed from this point.

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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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