i need help with a diffucult math question?

Question

who knows how to prove (2^n)(3^2n) -1 is always divisible by 17

Answer 1

(2^n)(3^2n) -1 = 18^n -1
18-1 = 17
18^2-1 = (18-1)(18+1) = 17 * 19
18^3 - 1 = (18-1)(18^2 + 1^2 + 18)
Go on like this. For a general term,
18^n - 1 = (18-1) * g(18,1) where g(18,1) is a polynomial involving 18 and 1
Clearly, it is divisible by (18 - 1) = 17

Answer 2

(2^n) (3^2n) -1 is always divisible by 17 becoz

2,3^2

2,3,3 are the least common multiples(factors) of 18 ie 2*3*3 = 18

and 17 = 18 -1. since 17 Is prime and doesnt have any numbers, its divisible to. u can only get its greatest multiples by using th factors of 18 minus 1. and thats why (2^n) (3^2n) -1 =is always divisible by 17

starting with the first number in th number line no. 1 If plugged in u get 17. 2*3^2 -1 =17. but this can only be achieved with minus 1 bcoz other wise u'll b gettin th factors of 17.

Answer 3

Let's see... based on ur question, we can conclude that (2^n)(3^2n) would be a multiple of 18. if u open up that, u get 2^n x (3^2)^n, which is equal to 9^n times 2^n and in turn equals to 18^n which is obviously an exponent of 18 and therefore is a multiple of 18. Proved.

Answer 4

(2^n)(3^2n) -1 = 18^n-1.

17 is a factor of 18^n-1. holds tru only for n as positive integer. so always divislble by 17

so (2^n)(3^2n) -1 is always divisible by 17 is NOT ture for Any n.

Answer 5

(2^n)(3^2n) -1
= (2^n)(9^n) -1
=18^n-1

here is a hint for you (i will not prove) :
(a^n - b^n) is divisible by (a-b)

in our case a=18, b=1
therefore this expression is divisible by 17

Answer 6

Use induction, it's all I really feel like giving you.
Since n is an integer.
Assume true for n=p for a general p and prove it's still true for n=p+1
Then substitute n=1 and verify divisibility of expression by 17.
Then you've proven the result by induction for all positive integers n.

Answer 7

2^n*3^(2n) -1
= 2^n*9^n -1
= (2*9)^n -1
= 18^n -1

let us take x^n - 1
put x = 1 so x^n-1 =0

x=1 is a zero for x^n-1
so (x-1) is a factor of x^n-1
so 17 is a factor of 18^n-1 putting x = 18

Answer 8

x = 2^n9^n - 1
x = 18^n - 1

since 18^n = 18^(n-1) X 18 = 18^(n-1)X(17+1)
= 18^(n-1)X17 + 18^(n-1)
also 18^(n-1) = 18^(n-2) X 18 = 18^(n-2)X(17+1)
= 18^(n-2)X17 + 18^(n-2)
until 18^(n-n) = 18

therefore:
18^n = 18^(n-1)X17 + 18^(n-2)X17 + 18^(n-3)X17 + ... + 18, for n > 0

x = 18^(n-1)X17 + 18^(n-2)X17 + 18^(n-3)X17 + ... + 18 - 1
x = 18^(n-1)X17 + 18^(n-2)X17 + 18^(n-3)X17 + ... + 17
x = 17 X (18^(n-1) + 18^(n-2) + ... + 1)

Therefore x is divisible by 17

Answer 9

You do it by getting off of the computer and doing the work. It won't get done all by itself. You need to figure this out because if you don't you will continue to be a village idiot.

Answer 10

mathematical induction