Question 1137739
Prove by induction that for all positive integers value of n:
{{{5^(2n)+3n-1}}} is an integer multiple of 9.<pre>
First we show that it is true when n=1:
{{{5^(2n)+3n-1}}}
{{{5^(2(1))+3(1)-1}}}
{{{5^2+3-1}}}
{{{25+3-1}}}
{{{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^(2n)+3n-1}}} equals to 9P.  That is, we examine
what would happen IF this were true for some value of k:

{{{5^(2k)+3k-1=9P}}}

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

{{{5^(2(k+1))+3(k+1)-1}}} that it will be a multiple of 9 also, so we simplify it:

{{{5^(2k+2)+3k+3-1}}}

{{{5^(2k)*5^2+3k+2}}}

{{{25*5^(2k)+3k+2}}}

We notice that the first term is 25 times the first term of {{{5^(2k)+3k-1}}}, 
so we see what we must add to 25 times {{{5^(2k)+3k-1}}} to get {{{25*5^(2k)+3k+2}}}.

25 times {{{5^(2k)+3k-1}}} is {{{25*5^(2k)+75k-25}}}. So we write

{{{25*5^(2k)+3k+2}}} as

{{{25*5^(2k)+(75-72)k-(25-27)}}} which is

{{{25*5^(2k)+75k-25-72k+27 }}}, which is

{{{25(5^(2k)+3k-1)-9(8k-3) }}}, which is

{{{25(9P)-9(8k-3) }}}, which is

{{{9(25P-8k+3)}}}, which is a multiple of 9.

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

{{{5^(2k)+3k-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^(2k)+3k-1}}} is a multiple of 9.

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

{{{5^(2k)+3k-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^(2k)+3k-1}}} is a multiple of 9.

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

{{{5^(2k)+3k-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^(2k)+3k-1}}} is a multiple of 9.

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

Edwin</pre>