Hello,Originally Posted byDenMac21

your proof looks OK to me. When I did this proof for my own (pleasure, of course!) I came to the same result. Here is my way to do it:

I use "complete induction" (I'm quite certain, that this isn't the correct expression. It's litterally translated from German. Sorry!). You have to use Peano's Axiom about .

1. A(n=1):

2. A(n): Let be true.

A(n+1):

All powrs of 6 have as last digit a 6, so the produkt 4*6^n has always a 4 as it's last digit. That means the number in the paranthesis has a 5 as last digit and is therefore divisible by 5. Therefore the 1rst summand is divisible by 25 too. Therefore the complete expression is divisible by 25.

Now you've proofed that your expression is true for (n+1) if it is true for n. That means: You've proofed, that it is true for n=1.

Because it is true for n=1 it is true for n=1+1=2 too;

because it is true for n=2 it is true for n=2+1=3 too;

because it is true for n=3 it is true for n=3+1=4 too.

So it is true for all

Maybe this was of some help for you.

Greetings

EB