Using the summation identity F₁ + F₂ + ... + Fₙ = Fₙ₊₂ − 1, what is F₁ + F₂ + F₃ + F₄ + F₅? (Recall: F₁=1, F₂=1, F₃=2, F₄=3, F₅=5, F₆=8, F₇=13)
A11
B12
C13
D14
By the identity, F₁+...+F₅ = F₇ − 1 = 13 − 1 = 12. You can verify directly: 1+1+2+3+5 = 12. The identity converts a five-term sum into a single lookup, which is precisely why these identities are useful — they compress calculation.
Question 2 Multiple Choice
Cassini's identity Fₙ₋₁Fₙ₊₁ − Fₙ² = (−1)ⁿ is most elegantly derived from which perspective?
ADirect computation for several values of n followed by strong induction
BApplying the Binet closed-form formula twice and simplifying
CRecognizing that det(Mⁿ) = (det M)ⁿ = (−1)ⁿ, where Mⁿ encodes Fibonacci numbers as entries
DCounting tiling arrangements of a strip of length n−1
The matrix M = [[1,1],[1,0]] has the property that Mⁿ = [[Fₙ₊₁, Fₙ],[Fₙ, Fₙ₋₁]]. The determinant of Mⁿ is (det M)ⁿ = (−1)ⁿ. Reading the determinant of the right-hand matrix directly gives Fₙ₊₁Fₙ₋₁ − Fₙ², so Cassini's identity follows immediately. This approach is most elegant because it reveals the identity as a consequence of matrix structure rather than a numerical curiosity.
Question 3 True / False
The general addition formula Fₘ₊ₙ = FₘFₙ₊₁ + Fₘ₋₁Fₙ reduces to the basic Fibonacci recurrence when m = 1.
TTrue
FFalse
Answer: True
Setting m = 1: F₁₊ₙ = F₁Fₙ₊₁ + F₀Fₙ. Since F₁ = 1 and F₀ = 0 (by convention), this gives Fₙ₊₁ = Fₙ₊₁, which is trivially the recurrence. More usefully, setting m = 2 gives F₂₊ₙ = F₂Fₙ₊₁ + F₁Fₙ = Fₙ₊₁ + Fₙ = Fₙ₊₂, which is exactly the definition. The general formula truly generalizes the recurrence.
Question 4 True / False
Cassini's identity shows that for any n, the product Fₙ₋₁Fₙ₊₁ equals Fₙ² exactly — consecutive Fibonacci numbers are perfectly correlated.
TTrue
FFalse
Answer: False
Cassini's identity states Fₙ₋₁Fₙ₊₁ − Fₙ² = (−1)ⁿ, which means the difference alternates between +1 and −1 — never zero. The flanking product is always exactly 1 away from the square, not equal to it. This near-miss is what makes the identity surprising and beautiful.
Question 5 Short Answer
Why is the matrix M = [[1,1],[1,0]] particularly powerful for deriving Fibonacci identities, rather than just using induction directly?
Think about your answer, then reveal below.
Model answer: Because Mⁿ encodes Fibonacci numbers as entries in a structured way, so identities about Fibonacci numbers become statements about matrix operations. Multiplying Mᵐ × Mⁿ = Mᵐ⁺ⁿ instantly yields the addition formula by reading off entries; taking the determinant of Mⁿ instantly yields Cassini's identity. The matrix framework generates many identities simultaneously from a single algebraic fact, rather than requiring a separate inductive proof for each.
Induction proves identities one at a time and requires guessing the right form in advance. The matrix approach provides a unified generating structure: since Mⁿ has a known form with Fibonacci entries, any algebraic property of matrices (determinant, trace, multiplication) automatically produces a Fibonacci identity. This is the deeper structural insight — the sequence is not just a list of numbers but the shadow of matrix exponentiation.