Prove by Induction that 3 raised to 2n+1 + 2 raised to n-1 is divisible by 7

[simcan18]

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?

 

[Deleted member 4993]

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?

You need to prove:

\(\displaystyle Prove \ that \ \displaystyle{3^{2n+1} \ + \ 2^{n-1}}\) is divisible by 7 for n=> 1

First step - show this is true for n=1 and continue.....

 

[simcan18]

Prove By Induction

[FONT=&quot]Some work that I did, but forgot to post. So, for the inductive step, a bit confused...not sure I'm heading in right direction


Base n = 1[/FONT]
[FONT=&quot] 3^2x1+1 + 2^1-1[/FONT]
[FONT=&quot] 3³ + 2⁰ = 27 + 1 =28 which is divisible by 7 so it is true for n=1[/FONT]
[FONT=&quot]
[/FONT]
[FONT=&quot] Assume true when n=k[/FONT]
[FONT=&quot] 3^2k+1 +2^k-1 = 7M, also could imply 3^2k+1 =7M-2^k-1[/FONT]
[FONT=&quot]
[/FONT]
[FONT=&quot] Inductive, So does it hold true for n=k+1[/FONT]
[FONT=&quot] 3^2(k+1)+1 +2^(k+1)-1 = 3^2k+2+1 +2^(k+1)-1[/FONT]
[FONT=&quot] =3^2k+1 x 3² + 2^k x 2⁰[/FONT]
[FONT=&quot] =9 x 3^2k+1 + 1 x 2^k[/FONT]
[FONT=&quot] = 27 x 3^2k +1x2^k

[/FONT]

 

[Dr.Peterson]

Some work that I did, but forgot to post. So, for the inductive step, a bit confused...not sure I'm heading in right direction


Base n = 1
3^2x1+1 + 2^1-1
3³ + 2⁰ = 27 + 1 =28 which is divisible by 7 so it is true for n=1


Assume true when n=k
3^2k+1 +2^k-1 = 7M, also could imply 3^2k+1 =7M-2^k-1


Inductive, So does it hold true for n=k+1
3^2(k+1)+1 +2^(k+1)-1 = 3^2k+2+1 +2^(k+1)-1
=3^2k+1 x 3² + 2^k x 2⁰
=9 x 3^2k+1 + 1 x 2^k
= 27 x 3^2k +1x2^k

I see that either you or someone else has been working on this same problem on my other site, and showed the same work. So if it is you, you may not need more help. But let me try, in case it isn't (and in case someone else is reading along here).

What you've done here under the inductive step is to take the expression that you need to show is divisible by 7 under the inductive hypothesis, and simplify it a bit. What you did there is valid, but not all helpful. Your goal is to rewrite the expression in a way that will somehow include that known multiple of 7, 3^2k+1 +2^k-1, so that you can conclude something. You nicely got it to include 3^2k+1 in your next to last line; you'd like to also include 2^k-1. So you might do this:

9 ⋅ 3^2k+1 + 1 ⋅ 2^k = 9 ⋅ 3^2k+1 + 2 ⋅ 2^k-1.

But that has the wrong coefficients, 9 and 2 rather than 1 and 1. Can you break up the first term into two terms (keeping in mind that we are looking for multiples of 7!) that will let you include the expression we want?