Question 1203186
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Translation of what the student wrote
<font color=blue>6x7^n — 2x3^n is divisible by 4 for every n natural numbers.
description: ^n = power n</font>


I'll write the solution in English, and let the student translate back to their original native language.


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I'm assuming that you want to prove this claim by induction.


The natural numbers is the set {1,2,3,4,...}
n = 1 is the smallest item
Compute the expression for n = 1
6*7^n - 2*3^n
6*7^1 - 2*3^1
6*7 - 2*3
42 - 6
36
4*9
The last step shows that we get a multiple of 4
Which in turn proves that 6*7^n - 2*3^n is a multiple of 4 when n = 1
In other words, 6*7^n - 2*3^n is divisible by 4 when n = 1


That covers the base case.
Now onto the inductive step.
Assume that 6*7^k - 2*3^k is a multiple of 4, when k is some natural number and k > 1.


This would mean 
6*7^k - 2*3^k = 4m
for some integer m


Based on the assumption above, we want to prove that  6*7^(k+1) - 2*3^(k+1) is also a multiple of 4
The goal is to show that 6*7^(k+1) - 2*3^(k+1) can be written in the template 4*(some integer).


Through a bit of algebra...
6*7^(k+1) - 2*3^(k+1)
6*7^k*7^1 - 2*3^k*3^1
42*7^k - 6*3^k
(6*7^k + 36*7^k) - (2*3^k + 4*3^k)
6*7^k + 36*7^k - 2*3^k - 4*3^k
(6*7^k - 2*3^k) + (36*7^k - 4*3^k)
(4m) + 4*(9*7^k - 1*3^k)
4( m + 9*7^k - 1*3^k )
4*(some integer)


We conclude that 6*7^(k+1) - 2*3^(k+1) is a multiple of 4 when 6*7^k - 2*3^k is a multiple of 4.
This wraps up the inductive step, and it wraps up the induction proof overall.



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Another approach to the inductive step


Return back to the assumption of 
6*7^k - 2*3^k = 4m


Isolating the first term gets us
6*7^k = 4m + 2*3^k


Then apply a substitution as marked in <font color=red>red</font>
6*7^(k+1) - 2*3^(k+1)
6*7^k*7^1 - 2*3^k*3^1
7*6*7^k - 6*3^k
7*<font color=red>6*7^k</font> - 6*3^k
7*(<font color=red>4m + 2*3^k</font>) - 6*3^k
28m + 14*3^k - 6*3^k
28m + 8*3^k
4*( 7m + 2*3^k )
4*(some integer)


We have shown that 6*7^(k+1) - 2*3^(k+1) is a multiple of 4 when 6*7^k - 2*3^k is a multiple of 4.


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However you approach the inductive step, both methods show that 6*7^k - 2*3^k being a multiple of 4 leads 6*7^(k+1) - 2*3^(k+1) to be a multiple of 4.


It is then a domino effect to cause 6*7^n - 2*3^n to be a multiple of 4 for any natural number n.
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