- #1

Anzas

- 87

- 0

prove that 3^(2n+1) + 4^(2n+1)

is divided by 7 for every natural n

is divided by 7 for every natural n

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- Thread starter Anzas
- Start date

In summary, we can prove that 3^(2n+1) + 4^(2n+1) is divisible by 7 for every natural n using induction and the property that c_1n + c_2m \equiv b_1n + b_2m\, (mod\, k) if c_1\equiv b_1,\,c_2\equiv b_2\,(mod\, k). We can also prove it without induction by rewriting the expression as a_n = 3^{2n+1}+4^{2n+1}\equiv 3^{2n+1} + (-3)^{2n+1} \equiv 3

- #1

Anzas

- 87

- 0

prove that 3^(2n+1) + 4^(2n+1)

is divided by 7 for every natural n

is divided by 7 for every natural n

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- #2

Zurtex

Science Advisor

Homework Helper

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What have you done so far?

- #3

snoble

- 127

- 0

So just write down your sequence [tex]a_n = 3^{2n+1}+4^{2n+1}[/tex]. When [tex]a_n[/tex] is in that form can you write down [tex]a_{n+1}[/tex] without out modifying the exponent.

Hope those hints help.

Steven

- #4

Anzas

- 87

- 0

im writing down what needs to be proven for n=n+1

3^(2n+3) + 4^(2n+3) / 7 = some integer = Z

9*3^(2n+1)+16*4^(2n+1) / 7 = Z

3^(2n+1)+4^(2n+1) / 7 + 8*3^(2n+1)+15*4^(2n+1) / 7 = Z

the first term is divisible because of the induction assumption so that leaves me with

8*3^(2n+1)+15*4^(2n+1) / 7

here I am stuck i can't find a way to show that this term is divisible by 7

i tried proving that with another induction but that led me no where.

i also tried transforming 8*3^(2n+1)+15*4^(2n+1) / 7 to

7*3^(2n+1)+7*4^(2n+1) / 7 + 3^(2n+1)+8*4^(2n+1) / 7

3^(2n+1)+4^(2n+1) + 3^(2n+1)+8*4^(2n+1) / 7

and that leaves me with the term 3^(2n+1)+8*4^(2n+1) / 7

that does not help me i can't find a way to show that its divisible by 7

any help/tips would be appreciated

sorry for not using latex by the way it would just take me too much time to type all this in latex

- #5

snoble

- 127

- 0

(3^(2n+1)+4^(2n+1)) / 7 + (8*3^(2n+1)+15*4^(2n+1)) / 7

= (3^(2n+1)+4^(2n+1)) / 7 + (3^(2n+1)+4^(2n+1)) / 7 + (7*3^(2n+1)+14*4^(2n+1)) / 7

You just needed to do your trick one more time over

Hope that helps

Steven

- #6

Anzas

- 87

- 0

of course! i can't believe i didn't notice this thank you very much!

- #7

Algebracus

- 2

- 0

[tex]a_n = 3^{2n+1}+4^{2n+1}\equiv 3^{2n+1} + (-3)^{2n+1} \equiv 3^{2n+1}\cdot(1 + (-1)^{2n+1}) \equiv 0 (mod 7)[/tex].

The most common method to prove this is by using mathematical induction. This involves proving that the statement is true for the base case (n=0) and then showing that if it is true for any value of n, it is also true for n+1. This can be done by manipulating the expression and using the fact that (a+b) divided by c is equal to (a/c + b/c).

The exponent (2n+1) represents an odd number, which is important because it allows us to use the property that (a+b)^n is always divisible by (a+b) for any positive integer n. This property can then be applied to both 3^(2n+1) and 4^(2n+1) in the expression.

Yes, there are other methods that can be used to prove this statement, such as direct proof or proof by contradiction. However, mathematical induction is the most commonly used method for this type of problem as it is systematic and straightforward.

Yes, by using mathematical induction, we can prove this statement for all positive integer values of n. This is because the inductive step shows that if the statement is true for any value of n, it is also true for n+1, thus covering all values of n.

Proving an expression is divisible by a specific number can have various applications in mathematics and science. It can be used to solve equations, simplify complicated expressions, or prove theorems. In this particular case, it can also be applied in number theory and cryptography.

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