Using complete induction, prove that if $a_1=2$, $a_2=4$, and $a_{n+2}=5a_{n+1}-6a_n$, then $a_n=2^n$ Could anyone please explain to me how to do this problem by using the principle of complete induction? Thanks. :)

Let $a_1=2$, $a_2=4$, and $a_{n+2}=5a_{n+1}-6a_n$ for all $n\geq 1$. Prove that $a_n=2^n$ for all natural numbers $n$.


For induction we wish to show the following:


*

*If $a_{n+2} = 2^{n+2}$ and $a_{n+1} = 2^{n+1}$ for some particular positive integer $n$ then: $a_{n+3} = 2^{n+3}$ must be true for that same value $n$

*$a_1 = 2^1$ and $a_2 = 2^2$
Since if 1 is true and 2 is true then we can repeatedly use 1 from the truth of 2 to establish that 


*

*$a_n = 2^n$ for all $n >= 1$


Proving point number 2 is trivial. It is already given to us. We now must prove point number 1.


*

*Assume that


$$a_{n+2} = 2^{n+2}$$
$$a_{n+1} = 2^{n+1}$$
For some arbitrary positive integer n.


*

*Notice that using the original definition:


$$a_{n+3} = 5*a_{n+2} - 6*a_{n+1} = 5*2^{n+2} - 6*2^{n+1}$$
Now we will simplify this expression:
$$5*2^{n+2} - 6*2^{n+1} =2*(5*2^{n+1})- (5*2^{n+1}) - 2^{n+1}$$
We now factor and reorganize:
$$2*(5*2^{n+1})- (5*2^{n+1}) - 2^{n+1} =(5*2^{n+1})(2 - 1) - 2^{n+1} = (5*2^{n+1}) - 2^{n+1} $$
And now simplify:
$$ (5*2^{n+1}) - 2^{n+1} = 4*2^{n+1} = 2*2^{n+2} = 2^{n+3} $$
Thus we have shown: that if 
$$a_{n+2} = 2^{n+2}$$
$$a_{n+1} = 2^{n+1}$$
and
$$a_{n+3} = 5*a_{n+2} - 6*a_{n+1}$$
Then:
$$a_{n+3} = 2^{n+3}$$
Thus we are now done with the induction.
