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Sample Test 4 Question 6

Let \(F_k=F_{k-1}+F_{k-2}\) where \(F_0=0, F_1=1\). Define the series \(s(x)= \sum_{k=1}^\infty F_k / x^k\) for \(x>2\).

Show that \(s(x)=\frac{x}{x^2-x-1}\).

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Hints

  • Hint 1
    Use the recursive definition of \(F_k\) to split \(s(x)\) into two sums. Carefully treat the base cases.
  • Hint 2
    Try rewriting the summation in terms of \(s(x).\)
  • Hint 3
    You may relabel the index of each sum in order to get it to match the definition of \(s(x).\)

Solution

The key is to breakdown \(F_k\) using its recursive definition, then relabelling the indices to get \(s(x)\) again. The first term of the sum \(\frac{F_1}{x}\) cannot be broken down as it is a base case, and must be taken out of the sum first. \[ \begin{equation} \begin{split} s(x)&=\sum_{k=1}^\infty \frac{F_k}{x^k}\\ &=\frac{F_1}{x} + \sum_{k=2}^\infty \frac{F_k}{x^k} \\ &=\frac{F_1}{x} + \sum_{k=2}^\infty \frac{F_{k-1}}{x^k} + \sum_{k=2}^\infty \frac{F_{k-2}}{x^k} \\ &=\frac{F_1}{x} + \sum_{i=1}^\infty \frac{F_{i}}{x^{i+1}} + \bigg(\frac{F_0}{x^2} + \sum_{j=1}^\infty \frac{F_{j}}{x^{j+2}}\bigg) \\ &=\frac{1}{x} + \sum_{i=1}^\infty \frac{F_{i}}{x^{i+1}} + \bigg(\frac{0}{x^2} + \sum_{j=1}^\infty \frac{F_{j}}{x^{j+2}}\bigg) \\ &=\frac{1}{x} + \frac{1}{x}\sum_{i=1}^\infty \frac{F_{i}}{x^i} + \frac{1}{x^2} \sum_{j=1}^\infty \frac{F_{j}}{x^j} \\ &=\frac{1}{x} + \frac{1}{x}s(x) + \frac{1}{x^2}s(x) \\ \end{split} \end{equation} \]

Finally, we rearrange to get \(s(x)=\frac{x}{x^2-x-1}\).

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