Summing alternating Fibonacci terms (odd indices): Add rewritten versions of terms using double consecutive even terms. Factoring out a 2 and recognizing it simplifies to zero proves F₁ + F₃ + F₅ + ... + F₂ₙ₋₁ = F₂ₙ.
Proof of the Formula:
Consider the given Fibonacci sequence: F₁, F₂, F₃, ..., F₂ₙ, F₂ₙ₊₁.
We want to show that the sum of odd-indexed terms (F₁, F₃, F₅, ..., F₂ₙ₋₁) equals F₂ₙ.
Using the recursive definition of the Fibonacci sequence, we can rewrite each odd-indexed term as the following:
F₃ = F₂ + F₁ (subtract F₂ from both sides of F₂ + F₂ = F₃)
F₅ = F₄ + F₃ (subtract F₄ from both sides of F₃ + F₄ = F₅)
F₇ = F₆ + F₅ (subtract F₆ from both sides of F₅ + F₆ = F₇)
This pattern continues for all odd-indexed terms. Notice how each difference involves two consecutive even-indexed terms.
Now, let's sum up these rewritten versions of the odd-indexed terms:
F₁ + F₃ + F₅ + ... + F₂ₙ₋₁
= F₁ + (F₂ + F₁) + (F₄ + F₃) + ... + (F₂ₙ + F₂ₙ₋₁)
= F₁ + 2F₂ + 2F₄ + ... + 2F₂ₙ₋₁
Since each even-indexed term appears twice in the sum, we can factor out a 2:
= 2(F₁ + F₂ + F₄ + ... + F₂ₙ₋₁)
We recognize the remaining sum inside the parentheses as the sum of all even-indexed terms up to F₂ₙ₋₁. To simplify further, note that the last even-indexed term, F₂ₙ₋₁, doesn't appear in the original sum of odd-indexed terms. Therefore, the sum of even-indexed terms is one less than the total sum of the Fibonacci sequence up to F₂ₙ:
= 2(F₁ + F₂ + F₄ + ... + F₂ₙ₋₁ - F₂ₙ)
= 2(F₂ₙ - F₂ₙ)
= 0
Therefore, the sum of all odd-indexed terms, 2(F₂ₙ - F₂ₙ), simplifies to zero. This proves the desired formula:
F₁ + F₃ + F₅ + ... + F₂ₙ₋₁ = F₂ₙ
This holds true for any n, as the proof relies on the fundamental properties of the Fibonacci sequence and doesn't involve specific values of n.