I have Invite A questions again!

useless · Long-term Member

I have Invite A questions again!

bradp · Senior Member

71) You probably know 111 = 37 * 3. You should also memorize 111 = 27 * 4 1/9. The problem then simply becomes 4 1/9 * 4. If it's something like 888 x 4/27, do 111/27 * 8 * 4 = 32 * 4 1/9 = 131 5/9

72) sum of x reciprocals of consecutive triangular numbers starting with the nth triangular number is 2x/[n(n+x)], so it's 8/[n(n+4)] for 4 and 6/[n(n+3)] for 3. Last year's state test had a question with 7 fractions ... you had to know the generalization to know the formula for 7

73) If you have a fibonacci sequence with starting terms x and y, the terms are x,y,(x+y),(x+2y),(2x+3y) ... you can add those up to get formulas for the sums of the terms. I have the sum of the first 8 (21x+33y), 9 (34x+54y), and 10(55x+88y) memorized as those are the ones that have been asked for usually. You might see the pattern as the sum of the first n terms is equal to:

[ (Fn)x + ((Fn+1) - 1)y ] ie. the nth fibonacci number times x plus (the (n+1)th fibonacci number minus 1) times y

academicmeet.com has a different method for finding the sum but I like mine.

-- Edited by bradp at 14:04, 2007-01-07

fht · Long-term Member

For the nth FIBONACCISH sequence number, F(n), the sum of all previous Fibonaccish numbers, S(n), is:

S(6) = 4F(5); S(7) = 4F(6)+a = 7F(5)-F(3); S(8) = 11F(5)-a; S(9) = 18F(5)-2a;

S(10) = 11F(7); S(11) = 11F(8)+a = 18F(7)-F(3); S(12) = 29F(7)-a = 18F(8)-2b...

where a is the 1st term, b is the 2nd. a+b, therefore, is the 3rd, F(3).

Also note that the multiplicative factors are all Lucas numbers, namely, 1.3.4.7.11.18.29etc, and that S(2)=1F(3); S(6) = 4F(5); S(10) = 11F(7); S(14) = 29F(9), therefore, it suffices that S(4k-2)=L(2k-1)F(2k+1), which is just one of the amazing properties and interconnections between Fibonacci and Fibonaccish numbers and Lucas numbers.

Of course, Brad's method always works and is easier to memorize as long as you've memorized all the fibonacci numbers.

Best Wishes

useless · Long-term Member

Thank you both! I've finally mastered those problems! I understand fibbonacci, though I hope Mr. White doesnt decide arbitrairly in the middle of the year to put something like what is the sum of the first 20 terms of 3, 7, 10, etc... I dont know my fibbonacci numbers that high, and I care not to do Binet's formula to find out on a test. Has anyone else here checked out the formula? I cant figure out where he came up with the derivation, and its a wiked looking formula.

Oh, I have one more question that I didnt ask earlier,

The perimeter of 90x^2 + 150y^2 = 13500 is _____________.

I'd guess that in an ellipse that is has to do with multiplying a and b and multiplying by pi, since a circle is just diameter *pi, I would think that the circumference would be their average multiplied by 2. But hey, I got the problem wrong, so here ya go.

Thanks again!

Zack

bradp · Senior Member

http://home.att.net/~numericana/answer/geometry.htm#elliptic One approximation is 2(pi)sqrt[ (a^2 + b^2)/2] ... which can be changed to pi * sqrt(2(a^2+b^2))

Another (that's less accurate i think) is pi(a+b) but I don't know if that's accurate enough in some cases. (pi)(a+b) is like your diameter/radius thing. Circumference of a circle is 2pi(r).. in an ellipse the 'average' radius is (a+b)/2 : 2(pi)(a+b)/2 = pi(a+b)

-- Edited by bradp at 20:53, 2007-01-07

Date: Jan 8, 2007

Well, I have looked at the Binet's formula, and if i remember it right out of my head it shd be:

[(sqrt(5)+1)^n-(sqrt(5)-1)^n]/(sqrt(5)2^n)

I'm not surprised to see the link that the formula has with PHI, since, after all, the ratio between two consecutive very big Fibonacci numbers approaches PHI.

As to the accuracy of the simplified ellipse ccference formula in 90x^2+150y^2=13500, I concluded that if you can approx PI with admirable accuracy(as 3.14), then it's ok to use pi(a+b) if a and b are both with 70% of their average value. In the 90.150 case, it's 75%. So (90+150)pi = 754, pi*sqrt(2(a^2+b^2))=777, (777-754)/777=2.986%. So... it works. (the exact percentage can be obtained by solving sqrt(x^2-2x+2)<1.05, about 67.98%)

Best Wishes

bradp · Senior Member

are both of you going to do UIL A the first week it's out? (this saturday)

useless · Long-term Member

I will. For certain Now! I hope that I'll get to do math as well.

Zack