Solution: In Exercises 129132, let an = (1 + 5) n (1 5) n 2n 5 be a sequence with nth
Chapter 9, Problem 132(choose chapter or problem)
In Exercises 129–132, let
\(a_{n}=\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n} \sqrt{5}}\)
be a sequence with nth term \(a_{n}\).
Use the result from Exercise 131 to show that \(a_{n+2}=a_{n+1}+a_{n}\). Is this result the same as your answer to Exercise 129? Explain.
Text Transcription:
a_n = (1 + sqrt 5)^n − (1 − sqrt 5)^n / 2^n sqrt 5
a_n
a_n+2 = a_n+1 + a_n
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