7 ms·

First one's simple enough, but second one I couldn't figure out if there's a clever trick. Or are students expected to apply the quadratic formula and do a bunc

by scarmig 13y ago

First one's simple enough, but second one I couldn't figure out if there's a clever trick. Or are students expected to apply the quadratic formula and do a bunch of symbolic manipulation?

asafira 13y ago
(Spoiler) Here's a way to do the second one: http://tinyurl.com/ojs6ux5 http://tinyurl.com/ojs6ux5
nilkn 13y ago
No factoring or quadratic formulas necessary. I'm going to write c_n instead of a_n and a,b instead of alpha,beta. Since a and b satisfy that equation, you know that a^2 = 6a+ 2, b^2 = 6b+ 2 and hence c_10 - 2 * c_8 = a^8 * (6a + 2) - b^8 * (6b + 2) + 2b^8 - 2a^8 = 6c_9. Dividing by 2c_9 gives 3. Edit: asafira beat me to it!
- alok-g 13y ago
  Using the same convention as yours, I tried to break c_10 in terms of c_9 as follows: (a^10-b^10) = (e a + f b) * (a^9-b^9) Multiplying out the RHS, e = f = 1 are the best choices, but there remained a residual. At a second glance though, this residual was proportional to (a^8-b^8). Problem solved.
Robin_Message 13y ago
I think approximation can be the best way to answer these types of multiple choice questions. Since its a quadratic, you know alpha is about 6 and beta is about -0.3. Then, since the answer is a whole number, and we're raising to high powers, you throw away the -0.3 and approximate a_n = 6^n. Divide through by a_8 and you get (6^2 - 2) / 12, which is very close to 3.
tzs 13y ago
There's an interesting connection between the roots of quadratic equations and Fibonacci-like sequences that is applicable to this problem. Let r and s be the roots of x^2 - ax - b, and let c[n] = (r^n - s^n)/(r - s). Then c[n] satisfies the recursion c[n] = a c[n-1] + b c[n-2]. For instance, if a = b = 1, then that sequence is the Fibonacci sequence, and the quadratic is x^2 - x - 1. r is then the golden ratio, s is -1/r, and (r^n - s^n)/(r-s) is the famous Binet formula for the n'th Fibonacci number. This applies directly to the ITT problem, giving the recurrence a_{n} = 6 a_{n-1} + 2 a_{n-2} for the a_{n} sequence from the problem, which in turn readily yields the answer.
- mjfl 13y ago
  I think it's easier to think about it as if you are just multiplying the original equations by some power of alpha or beta i.e. alpha^(n-2)(alpha^2 - 6alpha - 2) = 0 = alpha^n - 6 alpha^(n-1) - 2 alpha^(n-2) Hard to see without latex typing...
  - tzs 13y ago
    > Hard to see without latex typing... MathJax to the rescue: \[\alpha^{n-2}(\alpha^2-6\alpha-2)=0=\alpha^{n}-6\alpha^{n-1}-2\alpha^{n-2}\] To render with MathJax, use the bookmarklet available here: http://checkmyworking.com/misc/mathjax-bookmarklet/ http://checkmyworking.com/misc/mathjax-bookmarklet/

by scarmig 13y ago

First one's simple enough, but second one I couldn't figure out if there's a clever trick. Or are students expected to apply the quadratic formula and do a bunch of symbolic manipulation?

asafira 13y ago

(Spoiler) Here's a way to do the second one: http://tinyurl.com/ojs6ux5 http://tinyurl.com/ojs6ux5

nilkn 13y ago

No factoring or quadratic formulas necessary. I'm going to write c_n instead of a_n and a,b instead of alpha,beta. Since a and b satisfy that equation, you know that a^2 = 6a+ 2, b^2 = 6b+ 2 and hence c_10 - 2 * c_8 = a^8 * (6a + 2) - b^8 * (6b + 2) + 2b^8 - 2a^8 = 6c_9. Dividing by 2c_9 gives 3. Edit: asafira beat me to it!

alok-g 13y ago
Using the same convention as yours, I tried to break c_10 in terms of c_9 as follows: (a^10-b^10) = (e a + f b) * (a^9-b^9) Multiplying out the RHS, e = f = 1 are the best choices, but there remained a residual. At a second glance though, this residual was proportional to (a^8-b^8). Problem solved.

Robin_Message 13y ago

I think approximation can be the best way to answer these types of multiple choice questions. Since its a quadratic, you know alpha is about 6 and beta is about -0.3. Then, since the answer is a whole number, and we're raising to high powers, you throw away the -0.3 and approximate a_n = 6^n. Divide through by a_8 and you get (6^2 - 2) / 12, which is very close to 3.

tzs 13y ago

There's an interesting connection between the roots of quadratic equations and Fibonacci-like sequences that is applicable to this problem. Let r and s be the roots of x^2 - ax - b, and let c[n] = (r^n - s^n)/(r - s). Then c[n] satisfies the recursion c[n] = a c[n-1] + b c[n-2]. For instance, if a = b = 1, then that sequence is the Fibonacci sequence, and the quadratic is x^2 - x - 1. r is then the golden ratio, s is -1/r, and (r^n - s^n)/(r-s) is the famous Binet formula for the n'th Fibonacci number. This applies directly to the ITT problem, giving the recurrence a_{n} = 6 a_{n-1} + 2 a_{n-2} for the a_{n} sequence from the problem, which in turn readily yields the answer.

mjfl 13y ago
I think it's easier to think about it as if you are just multiplying the original equations by some power of alpha or beta i.e. alpha^(n-2)(alpha^2 - 6alpha - 2) = 0 = alpha^n - 6 alpha^(n-1) - 2 alpha^(n-2) Hard to see without latex typing...
- tzs 13y ago
  > Hard to see without latex typing... MathJax to the rescue: \[\alpha^{n-2}(\alpha^2-6\alpha-2)=0=\alpha^{n}-6\alpha^{n-1}-2\alpha^{n-2}\] To render with MathJax, use the bookmarklet available here: http://checkmyworking.com/misc/mathjax-bookmarklet/ http://checkmyworking.com/misc/mathjax-bookmarklet/