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**b.rad** Show that for all $\displaystyle n \geq 0, $

$\displaystyle 3^{2n}+4^{n+1}$ is a mulitple of 5 using an inductive proof.

I think that I have started this right but I am not sure if it is complete or correct.

$\displaystyle 1) \ p(0) = 3^{3(0)}+4^{(0)+1}=1+4=5$

$\displaystyle 2) \ Assume \ p(n)= 3^{2n}+4^{n+1} \ is \ multiple \ of \ 5,$

$\displaystyle p(n+1) = p(n)+3^{2(n+1)}+4^{(n+1)+1}$

$\displaystyle = 3^{2n}+4^{n+1}+3^{2n+2}+4^{n+2}$

$\displaystyle = 3^{2n}+4^{n+1}+3^{2n}(9)+4^{n+1}(4)$

$\displaystyle = 3^{2n}(1+9)+4^{n+1}(1+4) \ by \ factoring$

$\displaystyle = 3^{2n}(10)+4^{n+1}(5)$

So it appears to me that both terms are indeed multiples of 5; however, I thought the original p(n) would be factored out of the equation. In truth, it just looks a little funny. A little help or clarification would be greatly appreciated.