SOLUTION: Prove by induction that for all positive integers value of n: {{{5^(2n)+3n-1}}} is an integer multiple of 9.

Algebra ->  Sequences-and-series -> SOLUTION: Prove by induction that for all positive integers value of n: {{{5^(2n)+3n-1}}} is an integer multiple of 9.      Log On


   



Question 1137739: Prove by induction that for all positive integers value of n:
5%5E%282n%29%2B3n-1 is an integer multiple of 9.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Prove by induction that for all positive integers value of n:
5%5E%282n%29%2B3n-1 is an integer multiple of 9.
First we show that it is true when n=1:
5%5E%282n%29%2B3n-1
5%5E%282%281%29%29%2B3%281%29-1
5%5E2%2B3-1
25%2B3-1
27, 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, 5%5E%282n%29%2B3n-1 equals to 9P.  That is, we examine
what would happen IF this were true for some value of k:

5%5E%282k%29%2B3k-1=9P

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

5%5E%282%28k%2B1%29%29%2B3%28k%2B1%29-1 that it will be a multiple of 9 also, so we simplify it:

5%5E%282k%2B2%29%2B3k%2B3-1

5%5E%282k%29%2A5%5E2%2B3k%2B2

25%2A5%5E%282k%29%2B3k%2B2

We notice that the first term is 25 times the first term of 5%5E%282k%29%2B3k-1, 
so we see what we must add to 25 times 5%5E%282k%29%2B3k-1 to get 25%2A5%5E%282k%29%2B3k%2B2.

25 times 5%5E%282k%29%2B3k-1 is 25%2A5%5E%282k%29%2B75k-25. So we write

25%2A5%5E%282k%29%2B3k%2B2 as

25%2A5%5E%282k%29%2B%2875-72%29k-%2825-27%29 which is

25%2A5%5E%282k%29%2B75k-25-72k%2B27+, which is

25%285%5E%282k%29%2B3k-1%29-9%288k-3%29+, which is

25%289P%29-9%288k-3%29+, which is

9%2825P-8k%2B3%29, which is a multiple of 9.

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

5%5E%282k%29%2B3k-1 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 5%5E%282k%29%2B3k-1 is a multiple of 9.

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

5%5E%282k%29%2B3k-1 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 5%5E%282k%29%2B3k-1 is a multiple of 9.

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

5%5E%282k%29%2B3k-1 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 5%5E%282k%29%2B3k-1 is a multiple of 9.

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

Edwin