Question 170325
Use mathematical induction to prove: {{{4^(n+1) + 5^(2n-1)}}} is divisible by 21
<font size = 4 color = "red">Edwin's proof:
<pre><font size = 4 color = "indigo"><b>
Let {{{f(n)=4^(n+1) + 5^(2n-1)}}}

First prove that there is at least one
value of n for which {{{f(n)}}} is divisible by 21:

{{{4^(n+1) + 5^(2n-1)}}}
{{{4^(1+1) + 5^(2*1-1)}}}
{{{4^(2) + 5^(2-1)}}}
{{{16 + 5^(1)}}}
{{{16+5}}}
{{{21}}}

Strategy:
If n=k is a value of n so that f(n=k) is divisible by 21,
then if f(k+1) and f(k) differ by a multiple of 21, then 
f(k+1) will also be divisible by 21.

So, we will consider the difference f(k+1)-f(k)  

But first we must calculate f(n=k+1):

{{{f(n)=4^(n+1) + 5^(2n-1)}}}
{{{f(k+1)=4^((k+1)+1) + 5^(2(k+1)-1)}}}
{{{f(k+1)=4^(k+1+1)+5^(2k+2-1)}}}
{{{f(k+1)=4^(k+2)+5^(2k+1)}}}

Now we consider 

{{{ matrix(29,3,f(k+1)-f(k), "=", (4^(k+2)+5^(2k+1))-(4^(k+1) + 5^(2k-1)),
     "",     "=", 4^(k+2)+5^(2k+1)-4^(k+1) - 5^(2k-1),
          "",     "",  rearrange_terms,
     "",     "",   "",
     "",     "=", 4^(k+2)-4^(k+1)+5^(2k+1)-5^(2k-1),
     "",     "",   "",     
     "",     "",  factor_1st_2_terms_and_last_2_terms,
     "",     "",   "",
     "",     "=", 4^(k+1)(4-1)+5^(2k-1)(5^2-1),
     "",     "=", 4^(k+1)*3 +5^(2k-1)(25-1),
     "",     "=", 4^(k+1)*3 +5^(2k-1)*24,
     "",     "",   "",
     "",      "", write_24_as_3+21,
     "",      "",  "", 
     "",     "=", 4^(k+1)*3 +5^(2k-1)*(3+21),
     "",      "",   "",
     "",      "", distribute,
     "",      "",  "", 
     "",     "=", 4^(k+1)*3 +5^(2k-1)*3+5^(2k-1)*21,
     "",      "",   "",
     "",      "", factor_3_out_of_1st_2_terms,
     "",      "",   "",
     "",     "=", (4^(k+1)+5^(2k-1))*3+5^(2k-1)*21,  
     "",      "",   "",   
     "",     "=",  1st_parentheses_contains_f(k),
     "",      "",   "",
     "",     "=", f(k)+5^(2k-1)*21,
     "",     "",    "", 
     "",     "",  both_terms_are_divisible_by_21 )}}}

And since we know that since f(1) is divisible by 21
then f(2) is also divisible by 21, and thus f(3) is
also divisible by 21, etc., etc.

Edwin</pre>