Induction and the Fibonacci Sequence

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The discussion focuses on proving the identity $$\sum_{i=1}^n f^{2}{}_{i} = f_{n+1} * f_{n}$$ for the Fibonacci sequence defined as f1 = f2 = 1 and $$f_n = f_{n-1} + f_{n-2}$$ for n≥3. Strong induction is employed, with initial proofs for n=1 and n=2 confirming the identity holds. The proof progresses by assuming the hypothesis for k and attempting to prove it for k+1, where the simplification leads to recognizing that $$f_{k} + f_{k+1} = f_{k+2}$$. The conclusion is that the statement is proven, and it clarifies that the hypothesis does not need to assume k > 3, as the identity is valid for n ≥ 1.
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Homework Statement


Define the Fibonacci Sequence as follows: f1 = f2 = 1, and for n≥3, $$f_n = f_{n-1} + f_{n-2}, $$

Prove that $$\sum_{i=1}^n f^{2}{}_{i} = f_{n+1} * f_{n} $$

Homework Equations



See above.

The Attempt at a Solution



Due to two variables being present in both the Sequence and what needs to be proved, strong induction is required.

Proof that it works for n = 1 and n=2:

$$f_3 = f_{2} + f_{1} = 1 + 1 = 2 $$

n=1 LHS $$\sum_{i=1}^1 f^{2}{}_{i} = f^{2}{}_{1} = 1^2 = 1 $$
RHS $$f_{1+1} * f_{1} = 1 * 1 = 1$$

n=2 LHS $$\sum_{i=1}^2 f^{2}{}_{i} = f^{2}{}_{1} + f^{2}{}_{2} = 1^2 + 1^2 = 2 $$
RHS $$f_{2+1} * f_{2} = 2 * 1 = 2 $$

Hypothesis: $$\sum_{i=1}^k f^{2}{}_{k} = f_{k+1} * f_{k} $$

We want to prove: $$\sum_{i=1}^{k+1} f^{2}{}_{i} = f_{k+2} * f_{k+1} $$
It is in my proof where I do not quite know what I am doing:

$$\sum_{i=1}^{k+1} f^{2}{}_{i}
= \sum_{i=1}^{k} f^{2}{}_{i} + f^{2}{}_{k+1}
= f_{k+1} * f_{k} + f^{2}{}_{k+1}
= f_{k+1} (f_{k} + f_{k+1}) $$

How do I simplify from here? Not quite sure what to do with these functions.

Thanks all!
 
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t.kirschner99 said:

Homework Statement


Define the Fibonacci Sequence as follows: f1 = f2 = 1, and for n≥3, $$f_n = f_{n-1} + f_{n-2}, $$
Prove that $$\sum_{i=1}^n f^{2}{}_{i} = f_{n+1} * f_{n} $$

Homework Equations


See above.

The Attempt at a Solution


Due to two variables being present in both the Sequence and what needs to be proved, strong induction is required.

Proof that it works for n = 1 and n=2:$$f_3 = f_{2} + f_{1} = 1 + 1 = 2 $$
n=1 LHS $$\sum_{i=1}^1 f^{2}{}_{i} = f^{2}{}_{1} = 1^2 = 1 $$ RHS $$f_{1+1} * f_{1} = 1 * 1 = 1$$
n=2 LHS $$\sum_{i=1}^2 f^{2}{}_{i} = f^{2}{}_{1} + f^{2}{}_{2} = 1^2 + 1^2 = 2 $$ RHS $$f_{2+1} * f_{2} = 2 * 1 = 2 $$
Hypothesis: $$\sum_{i=1}^k f^{2}{}_{k} = f_{k+1} * f_{k} $$ We want to prove: $$\sum_{i=1}^{k+1} f^{2}{}_{i} = f_{k+2} * f_{k+1} $$
It is in my proof where I do not quite know what I am doing:$$\sum_{i=1}^{k+1} f^{2}{}_{i}
= \sum_{i=1}^{k} f^{2}{}_{i} + f^{2}{}_{k+1}
= f_{k+1} * f_{k} + f^{2}{}_{k+1}
= f_{k+1} (f_{k} + f_{k+1}) $$

How do I simplify from here? Not quite sure what to do with these functions.

Thanks all!
You're almost there !

Using the recursive definition of the Fibonacci sequence, what is ##\ f_{k} + f_{k+1} \ ? ##
 
SammyS said:
You're almost there !

Using the recursive definition of the Fibonacci sequence, what is ##\ f_{k} + f_{k+1} \ ? ##

Ah thank you for opening my eyes! That would be equal to $$f_{k+2} $$ thus the statement is proven.

One quick question about the hypothesis, would I need to assume that k > 3 ?
 
t.kirschner99 said:
Ah thank you for opening my eyes! That would be equal to $$f_{k+2} $$ thus the statement is proven.

One quick question about the hypothesis, would I need to assume that k > 3 ?
You're welcome!

At first I was also thinking that we should only look at k ≥ 3, but the thing you needed to prove was ##\ \displaystyle \sum_{i=1}^n f_{i}^{2} = f_{n+1} * f_{n}
\,.\ ## Everything in this equation is defined for n ≥ 1 .
 
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