# How do I prove that 2 x 34^n - 3x23^n + 1 is divisible by 726 for all positive integers n?I know you can use maths induction to prove this but I'm not sure how to solve induction questions that...

How do I prove that 2 x 34^n - 3x23^n + 1 is divisible by 726 for all positive integers n?

I know you can use maths induction to prove this but I'm not sure how to solve induction questions that have an unknown index.

### 1 Answer | Add Yours

`2 xx 34^n - 3xx23^n + 1`

When n = 1

`2 xx 34^n - 3xx23^n + 1 = 2 xx 34^1 - 3xx23^1+ 1 = 0`

0 is divisible by any number. So 0 is divisible by 726.

For n=1 the result is true.

Assume for n = p Where p is a positive integer the result is true.

So `2 xx 34^p - 3xx23^p + 1` is divisible by 726.

`2 xx 34^p - 3xx23^p + 1 = 726K` where `K in Z^+`

When n = p+1

`2 xx 34^(p+1) - 3xx23^(p+1) + 1`

`= 2 xx34^p*34^1 - 3xx23^p*23^1 + 1` since `a^(m+n) = a^m*a^n`

`= 34*2 xx 34^p – 23*3xx23^p + 1`

`= (23+11) 2 xx 34^p – 23*3xx23^p + 23-22`

`= 23*2 xx 34^p – 23*3xx23^p + 23-22+11*2 xx 34^p`

`= 23[2 xx 34^p - 3xx23^p + 1]+22(34^p-1)`

Now we have to prove `22(34^p-1)` is divisible by 726.

p=1

`22(34^p-1) = 22(34^1-1) = 22*33 = 726`

For p = 1 result is true.

At p=p assume that `22(34^p-1)` is divisible by 726.

`22(34^p-1) = 726A` Where A is a positive integer.

p=p+1

`22(34^(p+1)-1)`

`= 22*34^p*34-22`

`= 22*34^p*34-22*34-22+22*34`

`= 22*34(34^p-1)+22*34`

`= 34*726A+726`

`= 726[34A+1]`

`= 726*K1`

So for p=p+1 result is true.

So from mathematical induction for all positive integers p ; 22(34^p-1) is divisible by 726.

So ;

`2 xx 34^(p+1) - 3xx23^(p+1) + 1`

`= 23[2 xx 34^p - 3xx23^p + 1]+22(34^p-1)`

`= 23*726K+726K1`

`= 726[23K+K1]`

`= 726*K2`

K,K1 and K2 are all positive integers.

So for n=p+1 the result is true.

**So from mathematical induction for all positive integers n ; `2 xx 34^n - 3xx23^n + 1` is divisible by 726.**