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Thank you It's really not too difficult once it's explained, at least for polynomials.
I solved that system of equations and got a=b=0. I take that to mean that there is one solution of degree n-1 OR less?
Hi;
I worked out a formula for this information:
V = 1/3 * pi * √(L^2 * r^4 - r^6)
I think your reasoning is perfectly sound, and deducible from the rules given that
1. The number of even terms is the nth Fibonacci number
2. The number of odd terms is the n-1 Fibonacci number, and
3. The total number of terms is the n+1 Fibonacci number
For any given Sn.
Just to confirm, I will call a Fibonacci number that corresponds to the step number Fn, and restate exactly what you said in those terms.
Sn-1 has Fn-2 odds and Fn-1 evens.
Then Sn has Fn-1 odds and Fn = Fn-2 + Fn-1 evens,
The total number of elements being Fn+1 = Fn-2 + 2Fn-1
Sn+1 has Fn = Fn-2 + Fn-1 odds and Fn+1 = Fn-2 + 2Fn-1 evens
To slightly correct your running counts, it goes:
Odds: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34...
Evens: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...
Total: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89...
As far as I'm concerned, the only outstanding problem is how to obtain a general formula for the sum (there is also a little problem of obtaining a general formula for the smallest number, but that should be pretty easy). Everything else is toast by now, and it all has to do with Fibonacci.
That is trouble. A graphing calculator gave me an exponential that fits precisely as well. I suppose one rule will seem more implicit than another, but an open question shouldn't require a closed answer.
Wouldn't it be theoretically possible to construct several different (perhaps monstrously complex) formulas to fit the same sequence in many, if not most or all, cases?
A final note: A more intuitive way of solving 3. than just using the formula is to do what the "Prince of Mathematicians", Carl Gauss, apparently did as a schoolboy. He was told to write out the sum of natural numbers up to 100, and quickly solved it by recognising that most of the numbers could be matched up in pairs that add up to 100, for example 100+0=100, 99+1=100, 98+2=100... right down to 51+49 = 100, counting fifty pairs plus 50, giving a speedy answer of 5050.
The same can be done with sums up to 25, or any amount. With 25+0=25, 24+1=25, 23+2=25... you will promptly get to 13+12=25 and count thirteen pairs.
Hi Bob
I'm not sure whether there is supposed to be an additional term, but I don't think there is another term between 18 and 27 despite that double comma. There aren't many terms, and there might be several rules, but once again it looks like there is a fairly simple polynomial that will fit.
EDIT: No I think that is how it is supposed to be. The pattern is (9 + 3) + 6 + 9 + 12... The polynomial for that is
The sequence in 1. is not arithmetic or geometric; there is no common difference or common ratio. The pattern for 1. seems to be 1 + 2 + 3 + 4... These numbers are called triangular numbers, and I think the Pythagoreans revered them over 2500 years ago. There is a polynomial rule for this, it is
You can also use the above rule for Q3
For 2., you are correct.
Hi!
The inequalities are redundant.
Hey! I am very ignorant of geography even with so many clues, but I finally know one xD
Hello!
Hi;
It's a bit of a giveaway when you are given the first term
Hello;
How did you find school?
I wanted to be a philosopher............
Hello
Hello! I'm assuming ten years later is ten years in the future (i.e., later than now)
Hey! If I interpret correctly...
Hello!
Oh, I see Thank you, I was just completely unfamiliar with this type of problem and it took a while to see what was going on.
Are there any implicit assumptions about k and a? Do they have to be integers?
Oh... I knew there must be some hocus-pocus I was not privy to. Thank you for mentioning that