Well I believe that the answer is five. Simply because 10 divided by 5 is 2 and -15 divided by 5 is -3
Let Fn be the number of ways of arranging such flagpole with the given conditions.
When arranging a flagpole of n feet high, consider the following cases
If the last flag used is a red flag, then the other flags are n-1 foot high, so they can be seen as arranged on a smaller flagpole of n-1 feet high, which can be done in Fn-1 ways.
Similarly, If the last flag used is a gold flag, then the other flags can be seen as arranged on a smaller flagpole of n-1 feet high. This can be done in Fn-1 ways.
If the last flag used is green, the other flags are n-2 feet high, so the flagpole can be arranged in Fn-2 ways.
Using the sum rule, we obtain that Fn = 2Fn-1 + Fn-2 for all n≥3. Listing all the combinations of flags, the initial conditions are F1 = 2 and F2 = 3.
To Learn more about similar flagpole questions:
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A sequence can be generated by using an=4an−1 where a1=6 (a) 6, 20, 76, 300 (b)6, 24, 96, 384 (c)6, 10, 14, 18 (d) 6, 20, 100, 5
Tcecarenko [31]
Answer:
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- <u>Option (b) 6, 24, 96, 384 </u>
Explanation:
The complete question is:
A sequence can be generated by using 
where 
and n is a whole number greater than 1. What are the first four terms in the sequence.
The given rule is a recursive formula, which means that to find any term consecutive term, from the second one, you multiply the current term by 4.
Thus:
