The sum of , where each term is the -th Fibonacci number, shifted places to the right (that is, divided by ) can be represented as where Find .

First, let’s introduce some notation so that we have an easy way of talking about the terms here. Let be the -th Fibonacci number, i.e. , and . Let the sum we are supposed to evaluate be . Then

*(Note: If you are not comfortable with the summation signs, that’s ok! Just do everything in expanded sums with at the end. Summation signs are not that difficult but do take some getting used to; always try to substitute the 1st, 2nd and last value of the series to make sure that you got the summation limits and general term correct.)*

A very common technique for evaluating an infinite sum : **multiply by a suitable factor and try to make the “tails” (i.e. the part of the sum) match. Then, substitute on the right-hand side to get an equation in **. Most commonly, we will try to **manipulate the sum so that second term takes the place of the first term, and so on**.

Easiest to understand this through a worked example. In the problem at hand, we want to manipulate so that we can make into . The first thing we do is multiply by 10 so that the denominator matches:

Since , we have

To match the rest of the terms, we can use the Fibonacci identity :

In the last step, we simply used the definition of to substitute it back into the RHS. (*Isn’t it magical?*) Hence, we have a linear equation in each is easy to solve:

Hence, the answer to the problem is . The answer is **99**.

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