Prove that for all n ∈ N, prove that $ 3^{3n+1} + 2^{n+1} $ is divisible by 5.

So far what I've gotten is:

Step 1: We will prove this by using induction on n. Assume the claim is true when n = k. In other words, $$ 3^{3k+1} + 2^{k+1} $$

After that, I suppose I have to use k + 1?

  • 1
    $\begingroup$ Yeah don't forget that induction first step is actually n=1. step 2 is given n=k, prove n=k+1 $\endgroup$ – BCLC Oct 20 '14 at 18:53

Let for $n=k$ $5|(3^{3k+1}+2^{k+1})$
For $n=k+1$ we get $3^{3(k+1)+1}+2^{(k+1)+1}=3^{3k+1}*27+2^{k+1}*2=27*3^{3k+1}+27*2^{k+1}-27*2^{k+1}+2^{k+1}*2=27*(3^{3k+1}+2^{k+1})-27*2^{k+1}+2^{k+1}*2=27*(3^{3k+1}+2^{k+1})-25*2^{k+1}$
$5|(25*2^{k+1}) $ and $ 5|(3^{3k+1}+2^{k+1})$ so $ 5|(27*(3^{3k+1}+2^{k+1})-25*2^{k+1})$
You also need to show that for $n=1$:
$3^{3*1+1}+2^{1+1}=3^4+2^2=81+4=85$ and $5|85$
PS. If you don't understand something I can explain .

  • $\begingroup$ where does the 27 come from? $\endgroup$ – user3434743 Oct 20 '14 at 19:26
  • $\begingroup$ @user3434743 , $3^{3(k+1)+1}=3^{3k+4}=3^{3k+1}*3^3=3^{3k+1}*27$ $\endgroup$ – Antony Oct 20 '14 at 19:27

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