Math Is Fun Forum

SvenBee

Member

Registered: 2008-03-08

Posts: 106

Need help with this mathematical induction problem.

Prove that 2^(2n+1) + 3^(2n+1) is divisible by 5.

I completed the trivial first step and when n=1 the equation proves true.

So far I have:

f(n)=2^(2n+1) + 3^(2n+1), and since n=k and f(n)=f(k) then f(k) is divisible by 5 (or f(k)=5x, where x is an integer)

then:

2^(2k+1) + 3^(2k+1) for the second step

Inductive step:

2^(2(k+1)+1) + 3^(2(k+1)+1) = 2^(2k+3) + 3^(2k+3) = (4)2^(k+1)+(9)3^(k+1)

That's as far as I can get. I don't know what the next step is. PLEASE HELP!!!!

Thanks.

---

e...the red-headed stepchild of math.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Need help with this mathematical induction problem.

You're nearly there already, so I'll just give you a hint: 9 = 4+5.
Post again if you need a bigger nudge.

---

Why did the vector cross the road?
It wanted to be normal.

SvenBee

Member

Registered: 2008-03-08

Posts: 106

Re: Need help with this mathematical induction problem.

Thanks!

= (4)2^(2k+1)+(9)3^(2k+1)

= (4)2^(2k+1)+(4)3^(2k+1)+(5)3^(2k+1)

= (4)[2^(2k+1)+3^(2k+1)]+(5)3^(2k+1)

And since {2^(2k+1)+3^(2k+1)} is already divisible by 5 from the previous step and (5)3^(2k+1) is being multiplied by 5 it is logically divisible by 5, the entire f(k+1) is divisible by 5 and the proff is complete. qed.

Is that correct and thank you.

---

e...the red-headed stepchild of math.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Need help with this mathematical induction problem.

Yes that is correct. You should say that {2^(2k+1)+3^(2k+1)} is divisible by the assumption and not from the previous step. Remember that the whole point of induction is to prove that k+1 follows from k.