First we show that it is true when n=1:




, which is a multiple of 9 since 9∙3 = 27

Now we show that IF it were true when n=k, that it would also be true
when n=k+1.

We let a multiple of 9 be 9P where P is an integer such that
when n=k,  equals to 9P.  That is, we examine
what would happen IF this were true for some value of k:



we hope (but do not know!) that if that were true, then when we substitute
(k+1) for n, like this:

 that it will be a multiple of 9 also, so we simplify it:







We notice that the first term is 25 times the first term of , 
so we see what we must add to 25 times  to get .

25 times  is . So we write

 as

 which is

, which is

, which is

, which is

, which is a multiple of 9.

Now since we have a value of n, namely k=1, for which 

 is a multiple of 9, then we have

proved that we also have another value of n, namely k+1 or 1+1 or
2 for which  is a multiple of 9.

Now since we have a value of n, namely k=2, for which 

 is a multiple of 9, then we have

proved that we also have another value of n, namely k+1 or 2+1 or
3 for which  is a multiple of 9.

Now since we have a value of n, namely k=3, for which 

 is a multiple of 9, then we have

proved that we also have another value of n, namely k+1 or 3+1 or
4 for which  is a multiple of 9.

And so on and on for all positive integer values of n.

Edwin