Question

Let \[ \frac{z}{1 - z - z^2} = \sum_{n=0}^{\infty} a_n z^n, \quad a_n \in \mathbb{R} \] for all z in some neighbourhood of 0 in \(\mathbb{C}\).
Then the value of \(a_6 + a_5\) is equal to ................

Hint:

Recognizing common generating functions can save a lot of time. \(\frac{1}{1-z} = \sum z^n\), \(\frac{1}{(1-z)^2} = \sum (n+1)z^n\), and \(\frac{z}{1-z-z^2} = \sum F_n z^n\) are three very useful ones to remember.

Answer 1

Step 1: Understanding the Concept:
The given expression is the well-known generating function for the Fibonacci sequence. The coefficients \(a_n\) of the power series expansion are the Fibonacci numbers.
Step 2: Key Formula or Approach:
The generating function for the Fibonacci numbers \(\{F_n\}_{n=0}^{\infty}\), defined by \(F_0=0, F_1=1\) and \(F_{n} = F_{n-1} + F_{n-2}\) for \(n \ge 2\), is: \[ \sum_{n=0}^{\infty} F_n z^n = \frac{z}{1 - z - z^2} \] By comparing this with the given expression \(\sum_{n=0}^{\infty} a_n z^n = \frac{z}{1 - z - z^2}\), we can directly identify \(a_n = F_n\) for all \(n \ge 0\).
Step 3: Detailed Calculation:
The problem asks for the value of \(a_6 + a_5\). Based on our identification, this is equal to \(F_6 + F_5\).
Let's list the first few Fibonacci numbers:
- \(F_0 = 0\)
- \(F_1 = 1\)
- \(F_2 = F_1 + F_0 = 1 + 0 = 1\)
- \(F_3 = F_2 + F_1 = 1 + 1 = 2\)
- \(F_4 = F_3 + F_2 = 2 + 1 = 3\)
- \(F_5 = F_4 + F_3 = 3 + 2 = 5\)
- \(F_6 = F_5 + F_4 = 5 + 3 = 8\)
Now we can calculate the required sum: \[ a_6 + a_5 = F_6 + F_5 = 8 + 5 = 13 \] Alternatively, using the Fibonacci recurrence relation, \(F_n + F_{n+1} = F_{n+2}\). Therefore, \(F_5 + F_6 = F_7\).
Let's calculate \(F_7 = F_6 + F_5 = 8 + 5 = 13\).
The result is consistent.
Step 4: Final Answer:
The value of \(a_6 + a_5\) is 13.
Step 5: Why This is Correct:
The solution correctly identifies the power series coefficients as the Fibonacci numbers. The calculation of the 5th and 6th Fibonacci numbers and their sum is straightforward.