0
$\begingroup$

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?

$\endgroup$
  • 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
1
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.