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

  • MHB
  • Thread starter simcan18
  • Start date
  • Tags
    Induction
In summary, the conversation is about proving by induction that a given expression is divisible by 7 for all numbers greater than or equal to 1. The conversation includes discussing the base case, the inductive step, and rearranging the expression to involve 7.
  • #1
simcan18
6
0
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?
 
Mathematics news on Phys.org
  • #2
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
$3^{2(k+1)+1}+2^{(k+1)-1}$​

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

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

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

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.
 

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

What is "Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7"?

"Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7" is a mathematical statement that asks whether the expression 3^2n+1 + 2^n-1 is divisible by 7 for all positive integer values of n.

What is induction?

Induction is a mathematical proof technique that is used to prove statements for all positive integer values of n. It involves proving the base case (usually n = 1) and then assuming that the statement holds for some value of n and using that assumption to prove that it holds for the next value of n. This process is repeated until the statement is proven for all positive integer values of n.

How do you solve and prove by induction?

To solve and prove by induction, you must first prove the base case (usually n = 1). Then, you must assume that the statement is true for some value of n and use that assumption to prove that it is also true for the next value of n. This process is repeated until the statement is proven for all positive integer values of n.

How do you prove that an expression is divisible by a number?

To prove that an expression is divisible by a number, you must show that it can be divided evenly by that number without any remainder. This can be done by factoring the expression and showing that the number in question is a factor of every term in the expression.

Can you provide an example of a proof by induction?

Yes, an example of a proof by induction for the statement "3^2n+1 + 2^n-1 Divisible by 7" would involve proving the base case (n = 1) by showing that 3^3 + 2^0 is divisible by 7. Then, assuming that the statement is true for some value of n, it can be shown that it is also true for the next value of n by expanding the expression and using the fact that 7 is a factor of the previous term. This process can be repeated for all positive integer values of n, thus proving the statement by induction.

Similar threads

Replies
5
Views
2K
Replies
7
Views
3K
  • General Math
Replies
7
Views
2K
Replies
1
Views
672
  • Precalculus Mathematics Homework Help
Replies
10
Views
376
Replies
7
Views
2K
Replies
1
Views
856
  • Precalculus Mathematics Homework Help
Replies
11
Views
221
Replies
8
Views
2K
Back
Top