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.

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- Mar 12th 2008, 12:07 PMpizzaRusher1234induction
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. - Mar 12th 2008, 12:37 PMwingless
$\displaystyle f(n) = 3^{3n+1}+2^{n+1}$

$\displaystyle g(n)$ = 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 $\displaystyle n\geq 0$