## MMM #35 Fibonacci fun – we have winners!

I'm giving away three prizes this time. Yes, I did say I would give four prizes but when I reviewed the first two submissions one of them was not correct. Henno Brandsma, editor for the Topology Q+A Board, sent the first correct solution, and that was the only correct solution submitted before I gave a small hint. I'm giving a prize to .mau. since he has sent in many solutions to MMMs and has never been selected by random.org. And, I'm giving a prize to Olivier, who was selected by random.org.

Here was the problem:

Let F(0)=1, F(1)=1, and F(n)=F(n-2)+F(n-1). This is the familiar Fibonacci series.

Simplify F(0)xF(1) + F(1)xF(2) + F(2)xF(3) + F(3)xF(4) + … + F(n-1)xF(n) + F(n)xF(n+1)

Show your work.

I got 14 submissions. Most of you sent induction proofs. Since induction doesn't give any insight as to how a solution was derived I can only guess that folks found patterns for sums as n increases then used induction to verify the patterns they saw.

Chao Xu submitted a nice induction proof at the Math Solutions Blog. The solution was password protected to not help others before the submission deadline. I've asked Chao to unprotect it.

Duram had an interesting different-looking but equivalent answer:

S(n+1)=( f(n+2)^2 - f(n) x f(n+1) -1 ) /2

Note how this answer eliminates the need to consider odd and even cases. And, yes, the index is off by 1 in this formula but the insight is excellent.

Jacques Descartes had a clever simplification of the odd/even thing:

S(n)=F(n+1)^2-.5+.5(-1)^n

Do you see how odd/even is handled?

Matthias Malandain proved these two formulas:

S(2n) = (F(2n+1))^2

S(2n-1) = F(2n-1) x F(2n+1)

Another way to approach the problem was by looking at the pictures I posted. While my way does not provide a proof it gives one an intuitive sense of why the answer is what it is.

Watchmath submitted a nice proof that didn't use recursion.

I'm delighted to see the variety in solving this problem.

Stay tuned Monday for the next MMM, right here at Wild About Math!

LarryJuly 4th, 2009 - 12:31

Hey I was just looking for somewhere to post this…this problem/solution is part of the practice for the GED at a website…it seems totally wrong…here is what I sent to the administrator in email…am i wrong???

Emailed this:

I am truly confused by this problem and its solution….it seems totally incorrect. Please email me back with where I am going wrong?

It is located in the Math Section 1B

Problem – Proportions:

At the market, a man buys six avocados for $13.35. All produce is 30% off today only. How much will two avocados cost you tomorrow at full price?

A) $4.00

B) $4.50

C) $5.00

D) $5.50

E) $6.00

You want to find the regular price per avocado because the question is asking you to calculate the cost of two of them.

You can find this by figuring out the sale price. The man paid $13.35 for 6 avocados. So divide the total cost by the number of avocados to get the price for one avocado.

13.35/6 = 2.225 this is the discounted price per avocado.

The discount is 30% less than the regular price.

If you take away 30%, you have 70% (since 100% – 30% = 70%)

So you know that $2.225 is 70% of the full price.

Full $ x 0.70 = 2.225

(Remember that if A x B = C, then A = C /B)

So full $ = 2.225 / 0.70 which, when we work it out either with the calculator or by hand, gives us = $ 2.50

At $2.50 per avocado, buying two would cost us $5.00, which is Answer C.

I WROTE:

I am no math wiz, but If I break this down to a simpler form:

If you say that the full price of a avocado is 2.50 and you discounted it 30% = 2.50 x .30 = .75 then 2.50 – .75 = 1.50

Wouldn’t the discounted price (from your solution) be 1.50 per avocado, not the original discounted price of 2.225?

In short you are saying that 2.22 is 30% off of 2.50???

I come up with 3.17 as the full price per avocado as 2.22 is 30% off of 3.17, in which the answer would be 3.17 x 2 = 6.34