Math Olympiad | Do you have a mathematical intuition?
ฝัง
- เผยแพร่เมื่อ 27 ส.ค. 2024
- A great question. If you're reading this ❤️.
What do you think about this problem?
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
Let f(n)=a^n+b^n. Given f(1)=1 and f(2)=2. f(n+2)= f(1)*f(n+1) - ((f(1)^2 - f(2))/2)*f(n) = f(n+1) + f(n)/2. Then f(3) = 5/2, f(4) = 7/2,... f(8)=97/8. Also (f(n+1)/f(n) converges to (f(1) + sqrt(2*f(2) - f(1)^2))/2 = (1+sqrt(3))/2)
(a + b)² = 1 = a² + b² + 2ab = 2 + 2ab
ab = -1/2
(a² + b²)² = 4 = a⁴ + b⁴ + 2(ab)² = 4
a⁴ + b⁴ = 4 - 1/2 = 7/2
(a⁴ + b⁴)² = a⁸ + b⁸ + 2(ab)⁴ = 49/4
a⁸ + b⁸ = 49/4 - 1/8 = 97/8
Basic Algebra. Good work...
Let A_n = a^n + b^n
A_1 = 1
A_2 = 2
recursion formula is...
A_n+1 = A_n + (1/2)A_n-1
hence, general term is...
A_n = [{(3+√3)/2}{1+√3}/2]^(n-1)
- {(3-√3)/2}{1-√3}/2]^(n-1)]/√3
Nope.
I used the first two equation to get a and b. a=1-b => b²+(1-b)²=2 => quadratic equation => a and b. But this takes a bit longer. since then you have to get 1/2 +/- 1/2*SQRT(3) to the 4th power which is time consuming, although not complicated mathematically.
We know a & b are interchangeable, so just evaluate with a ≥ b. To go from a & b to a⁸ & b⁸, just do square & simplify for each variable 3 times, and finally add them.
We know: a + b = 1 [eq. 1]; a² + b² = 2 [eq. 2]. Find: a⁸ + b⁸ .
We see a & b are interchangeable, so we only need to get one solution for a & b, if at all.
Square eq. 1, substitute with eq. 2, simplify:
a² + 2*a*b + b² = 1
2 + 2*a*b = 1
a*b = -1/2 [eq. 3]
Square eq. 2, substitute with eq. 3, simplify:
a⁴ + 2*a²*b² + b⁴ = 4
a⁴ + 1/2 + b⁴ = 4
a⁴ + b⁴ = 7/2 [eq. 4]
Square eq. 4, substitute with eq. 3, simplify:
a⁸ + 2*a⁴*b⁴ + b⁸ = 49/4
a⁸ + 1/8 + b⁸ = 49/4
a⁸ + b⁸ = 97/8
Alternatively, we solve for a & b, and compute. Multiply eq. 1 by a, substitute eq. 3, solve for a:
a² + a*b = a
a² - 1/2 - a = 0
2*a² - 2*a - 1 = 0
a & b are interchangeable. Let a ≥ b, to work once.
a = (1 + √3)/2, b = (1 - √3)/2
Square a, b 3 times and add:
a² = (2 + √3)/2, b² = (2 - √3)/2
a⁴ = (7 + 4*√3)/4, b⁴ = (7 - 4*√3)/4
a⁸ = (97 + 56*√3)/16, b⁸ = (97 - 56*√3)/16
a⁸ + b⁸ = 97/8
(b-1)*2+b*2=2
a=1-b
Square a+b to find ab=-1/2. Then square a^2+b^2 to give a^4+b^4, and again to give a^8+b^8 = 97/8. What is the problem?
Very complicatie answer: just do a2, b2, then a4, b4 and then a8, b8 and you get 97/8 in less that 5 minutes
Amazing…how to train properly to improve my int at this level?
❤
Answer 8 because a+b=1,a2+b2=2 so ab =1/2. (a^2+b^2)^4-3ab(a+b)-6(ab)^2=a^8+b^8
Nope.
Excellent
a² + (a - 1)² = 2
2a² - 2a + 1 = 2
2a² - 2a - 1 = 0
a = (2 ± √(4 + 8))/4
a = 1/2 ± √12/4
a = 1/2 ± √3/2
Let a = 1/2 + √3/2, b = 1/2 - √3/2
a² = 1/4 + √3/2 + 3/4, b² = 1/4 - √3/2 + 3/4
a² = 1 + √3/2, b² = 1 - √3/2
a⁴ = 1 + √3 + 3/4, b⁴ = 1 - √3 + 3/4
a⁴ = 7/4 + √3, b⁴ = 7/4 - √3
a⁸ = 49/16 + 7√3/2 + 3
b⁸ = 49/16 - 7√3/2 + 3
a⁸ + b⁸ = 49/8 + 6 = 97/8