# Thread: Proof by induction

1. ## Proof by induction

Hello guys could anyone help find a solution to this problem,
Show that:

3 x 5^(2n-1) + 2^(3n-2) is divisble by 17 (for every n on N*)

This is what i did :
verification for n = 1.
then we assume it is divisible for n and show it is also divisble for n+1

3 x 5^(2n+2-1) + 2^(3n+3-2)
= 3 x 5^(2n+1) + 2^(3n+1)
= 3 x 5^(2n-1) x 5^2 + 2^(3n-2) x 2^3
i tried a lot of things but i couldnt find the solutions.

2. ## Re: Proof by induction

Let the statement be true for n = k
That means 3 * 5^(2k-1) + 2^(3k-2) = 17 p for some p in N
That gives 3 * 5^(2k-1) = 17 p - 2^(3k-2) ------- [ 1 ]
Now for n = k+1 we have
3 * 5^{2(k+1) -1} + 2^{3(k+1)-2} = 3 * 5^(2k+1) + 2^(3k+1) = 3 * 5^(2k-1+2) + 2^(3k-2+3)=3 * 25 *5^(2k-1) + 8*2^(3k-2)
= 25* { 17 p - 2^(3k-2)} + 8*2^(3k-2) = 25* 17 p - 25* 2^(3k-2) + 8*2^(3k-2)= 25* 17 p - 17* 2^(3k-2)
And you have the result

3. ## Re: Proof by induction

Originally Posted by ibdutt
Let the statement be true for n = k
That means 3 * 5^(2k-1) + 2^(3k-2) = 17 p for some p in N
That gives 3 * 5^(2k-1) = 17 p - 2^(3k-2) ------- [ 1 ]
Now for n = k+1 we have
3 * 5^{2(k+1) -1} + 2^{3(k+1)-2} = 3 * 5^(2k+1) + 2^(3k+1) = 3 * 5^(2k-1+2) + 2^(3k-2+3)=3 * 25 *5^(2k-1) + 8*2^(3k-2)
= 25* { 17 p - 2^(3k-2)} + 8*2^(3k-2) = 25* 17 p - 25* 2^(3k-2) + 8*2^(3k-2)= 25* 17 p - 17* 2^(3k-2)
And you have the result
Thanks.

4. ## Re: Proof by induction

How to make this post solved ?