Ugly Fibonacci Sums

Algebra Level 4

If $k$ is an integer that satisfies the identity $\left(\sum_{i=1}^{n}F_{2i-1}\right)^2-\left(2\sum_{i=1}^{n}F_{2i-1}^2\right)=\left(\sum_{i=1}^{2n-2}(-1)^iF_i^2\right)-k$ then find $k$ .

$\text{Details and Assumptions}$

$F_1=F_2=1$ , and $F_n=F_{n-1}+F_{n-2}$ .

[You should assume that $n \geq 2$ for the summations to make sense.]

The answer is 1.

1 solution

Finn Hulse
Apr 26, 2014

Well, I'm sure there's a really neat, beautiful approach to this, but I just bashed it out.

Armed with the knowledge that this is indeed an identity, I assumed $n=2$ . Thus, I attained $(1+2)^2-2((1)^2+(2)^2)=0-k$ by evaluating each summation. Upon solving, I found $k=\boxed{1}$ . I'll be expecting an epic proof from @Daniel Liu , though. :D

Okay, the intended solution is to just plug a number in and solve. This problem was designed to be a troll problem :P

However, there is a very clever Proof with No Words. I will give you the picture but it is up to you to figure out why it works.

Imgur Imgur

Daniel Liu - 7 years, 1 month ago

OH MY GOD I JUST REALIZED THAT HOLY CRAP THAT'S THE COOLEST THING I'VE EVER SEEN OH MY GOD HOLY CRAP! THAT'S SO EPIC IT ALL MAKES SENSE NOW! :D

Finn Hulse - 7 years, 1 month ago

I assumed $n=1$ . (The right hand sum is simply zero.) :P

Ivan Koswara - 7 years, 1 month ago

Yeah, but then what about the second half where it's the sum from $1$ to $2n-1$ ? Wouldn't that translate to be $1$ to $0$ with $0$ being the upper limit? That's impossible, there is no $0$ th Fibonacci number. Did you assume it would be 1? Because then, you were REALLY lucky. :D

Finn Hulse - 7 years, 1 month ago

A sum for indices $1$ to $0$ is simply an empty sum.

Also, $F_{-1} = F_1 - F_0 = 1$ , by extending the identity $F_{n+2} - F_{n+1} = F_n$ backwards.

Ivan Koswara - 7 years, 1 month ago

@Ivan Koswara – When the indices are flipped (in reverse order), you actually start taking negative sums. One way to interpret it, is to consider the identity:

$\sum_{i=a}^b x_i + \sum_{i=b+1}^{c} x_i = \sum _{i=a}^c x_i.$

Set $a = 1, b= 0, c = 1$ , and we get that

$\sum_{i=1}^0 x_i + \sum_{i=1}^{1} x_i = \sum _{i=1}^1 x_i \Rightarrow \sum_{i=1}^{0} x_i = 0 .$

This is similar to the argument that $\int_a^b = - \int_b^a$ .

Calvin Lin Staff - 7 years, 1 month ago

@Ivan Koswara – Ah, true true. I guess you could extend the Fibonacci numbers all the way back to $F_{-\infty}$ .

Finn Hulse - 7 years, 1 month ago

I tried to do that but I guess I added wrong and didn't check my work so I was like "what the heck how am I wrong" so yeah I just revealed solutions

Nathan Ramesh - 7 years, 1 month ago

I hate when I do all this hard wok and in the end do my arithmetic wrong. It pisses me off. D:

Finn Hulse - 7 years, 1 month ago

Ugly Fibonacci Sums

The answer is 1.

1 solution

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