The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. 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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