prove
f_{2n+1} = f_{n+1}^2 + f_n^2
Where f is the fibonacci sequence (so f_1=1, f_2=1, f_3=2, f_4=3, f_5=5, etc)
To prove the equation f_{2n+1} = f_{n+1}^2 + f_n^2, we can use the mathematical principle of mathematical induction. Let's first define the Fibonacci sequence:
f_1 = 1
f_2 = 1
f_{n+2} = f_{n+1} + f_n (for n ≥ 1)
We want to show that for any positive integer n, the equation f_{2n+1} = f_{n+1}^2 + f_n^2 holds true.
Step 1: Base Case
Let's check if the formula holds true for the base case of n = 1:
For n = 1:
f_{2(1)+1} = f_3 = 2
f_{1+1}^2 + f_1^2 = f_2^2 + f_1^2 = 1^2 + 1^2 = 2
The equation holds true for the base case.
Step 2: Inductive Hypothesis
Assume that for some positive integer k, the equation holds true:
f_{2k+1} = f_{k+1}^2 + f_k^2
This is our inductive hypothesis.
Step 3: Inductive Step
We need to show that if the equation holds true for k, it also holds true for k+1.
For n = k+1:
f_{2(k+1)+1} = f_{2k+3}
We can rewrite this as:
f_{2k+1+2} = f_{2k+1+1} + f_{2k+1}
Using the formula for the Fibonacci sequence:
f_{2k+1+2} = f_{2k+2} + f_{2k+1}
Applying the inductive hypothesis:
f_{2k+1+2} = (f_{k+1+1})^2 + (f_{k+1})^2 + f_{2k+1}
Simplifying, we get:
f_{2k+1+2} = f_{k+2}^2 + f_{k+1}^2 + f_{2k+1}
Now, let's express f_{2k+1} in terms of f_k and f_{k+1} using the Fibonacci sequence formula:
f_{2k+1} = f_{k+1}^2 + f_k^2
Substituting this into our equation, we have:
f_{2k+1+2} = f_{k+2}^2 + f_{k+1}^2 + (f_{k+1}^2 + f_k^2)
Simplifying, we get:
f_{2k+1+2} = f_{k+2}^2 + 2f_{k+1}^2 + f_k^2
Now, let's express f_{k+3} in terms of f_{k+1} and f_k using the Fibonacci sequence formula:
f_{k+3} = f_{k+2} + f_{k+1}
Substituting this into our equation, we have:
f_{2k+1+2} = (f_{k+2} + f_{k+1})^2 + 2f_{k+1}^2 + f_k^2
Expanding and simplifying, we get:
f_{2k+1+2} = f_{k+2}^2 + 2f_{k+2}f_{k+1} + f_{k+1}^2 + 2f_{k+1}^2 + f_k^2
Combining like terms, we have:
f_{2k+1+2} = f_{k+2}^2 + 3f_{k+1}^2 + f_k^2 + 2f_{k+2}f_{k+1}
Rearranging the terms, we get:
f_{2k+1+2} = f_{k+2}^2 + f_{k+1}^2 + (f_{k+1}^2 + f_k^2) + 2f_{k+2}f_{k+1}
We can recognize that the terms in parentheses are f_{k+1}^2 + f_k^2, which is our inductive hypothesis.
Therefore,
f_{2k+1+2} = f_{k+2}^2 + f_{k+1}^2 + f_{k+1}^2 + f_k^2 + 2f_{k+2}f_{k+1}
= f_{k+2}^2 + 2f_{k+1}^2 + f_k^2 + 2f_{k+2}f_{k+1}
This equation is equivalent to our original equation for n = k+1.
Since the equation holds true for the base case and for k+1 whenever it holds true for k, we conclude that the equation f_{2n+1} = f_{n+1}^2 + f_n^2 is true for all positive integers n, by mathematical induction.