induction

**pizzaRusher1234** · March 12th 2008, 12:07 PM

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.

**wingless** · March 12th 2008, 12:37 PM

$f(n) = 3^{3n+1}+2^{n+1}$

$g(n)$ = Remainder of $\frac{f(n)}{5}$

We'll prove that $g(n) = 0$ for all $n$ integers.

Firstly, try it for an integer, for example $n=0$ ..
$f(0) = 3^{1}+2^{1} = 5$
$g(0) = 0$
It satisfies.

Now we can use induction.

$f(n+1) = 3^{3n+4}+2^{n+2}$
$f(n+1) = f(n) + 26\cdot 3^{3n+1} + 2^{n+1}$
$f(n+1) = f(n) + \underbrace{25\cdot 3^{3n+1}}_{\text{Multiple of 5}} + \underbrace{3^{3n+1} + 2^{n+1}}_{\text{Multiple of 5}}$

So if f(n) is divisible, f(n+1) is divisible too.

Note: This doesn't work for all n integers. It works for $n\geq 0$

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