## Fibbo-something sequences

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
4mase
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### Fibbo-something sequences

Does anybody understand the concept of analyzing a word problem and turning it into an equation?

EX

Mr. Yazaki discovered that there were 225 dandelions in his garden on the first Saturday of spring. He had time to pull out 100, but by the next Saturday, there were twice as many as he had left. Each Saturday in spring he removed 100 dandelions, only to find that the number of remaining danelions had doubled the following Saturday.

So can somebody explain how to solve this?

stapel_eliz
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Does anybody understand the concept of analyzing a word problem and turning it into an equation?
In general, sure. But the techniques, etc, are covered in various courses, generally over a period of years.
Mr. Yazaki discovered that there were 225 dandelions in his garden on the first Saturday of spring. He had time to pull out 100, but by the next Saturday, there were twice as many as he had left. Each Saturday in spring he removed 100 dandelions, only to find that the number of remaining danelions had doubled the following Saturday.

So can somebody explain how to solve this?
Um... So far, all I can see is a story. I see no equation to "solve", or even any instructions on what to do with this...?

Eliz.

little_dragon
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### Re: Fibbo-something sequences

I think the "Fibbo-something" is "Fibonacci", like "fib a knot chee". I think you start with 1 and 1, and then you get your next numbers by adding the two before. Maybe you're supposed to do something like that, and come up with some kind of formula for how many you have each Sunday.

Karl
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### Re: Fibbo-something sequences

The problem appears to be asking for the sequence of the number of dandelions each Saturday. The problem gives that on the first Saturday, the number of dandelions is 225. So
`n[0] = 225`
To get from any Saturday's number to the next Saturdays number, you subtract 100 and double the result. The equation for that is
`n[k+1] = 2(n[k]-100)`
The above is called a recursion equation, because it tells you how to get the next in the sequence from the last value or values. Crunching the numbers for the first few in the sequence, I get
`n[1] = 250`
`n[2] = 300`
`n[3] = 400`
The real question is, can you find a closed expression for n[k]? That is, if I tell you which member of the sequence I want to know the value of, is there a way, without calculating all the n's up to that point, that you can compute for me the value of that member of the sequence. Look at the following table of expanding the calculation of the first few values, and you can see a pattern:
`n[0] = 225`
`n[1] = 21(225) - 21(100)`
`n[2] = 22(225) - 22(100) - 21(100)`
`n[3] = 23(225) - 23(100) - 22(100) - 21(100)`
In general, for the k'th value, you take 225 times 2k and subtract the sum of all the powers of 2 from there down times 100. The sum of powers of 2 from 1 to k is 2k+1 - 2. So the general formula for the k'th value is
`n[k] = 2k(225) - 100(2k+1 - 2)`
I think this is the kind of work the problem was asking for.

Challenge: Can you prove that the above formula is equivalent to
`n[k] = 200 + 2k(25)`