Hi,
We're on induction right now. Can somebody help me?
Prove that for all integers n, 3^(3n+1) + 2^(n+1) is a multiple of 5.
$\displaystyle f(n) = 3^{3n+1}+2^{n+1}$
$\displaystyle g(n)$ = Remainder of $\displaystyle \frac{f(n)}{5}$
We'll prove that $\displaystyle g(n) = 0$ for all $\displaystyle n$ integers.
Firstly, try it for an integer, for example $\displaystyle n=0$..
$\displaystyle f(0) = 3^{1}+2^{1} = 5$
$\displaystyle g(0) = 0$
It satisfies.
Now we can use induction.
$\displaystyle f(n+1) = 3^{3n+4}+2^{n+2}$
$\displaystyle f(n+1) = f(n) + 26\cdot 3^{3n+1} + 2^{n+1}$
$\displaystyle f(n+1) = f(n) + \underbrace{25\cdot 3^{3n+1}}_{\text{Multiple of 5}} + \underbrace{3^{3n+1} + 2^{n+1}}_{\text{Multiple of 5}}$
So if f(n) is divisible, f(n+1) is divisible too.
Note: This doesn't work for all n integers. It works for $\displaystyle n\geq 0$