The characteristic polynomial for this equation is:
It certainly is a lot faster and more elegant, but when you say it isn't taught in schools, does that mean we can't actually use it in the HSC?PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).
Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.
I'll summarise my method for the previous question in case I was ambiguous:
1. Make the characteristic polynomial.
2. Solve the characteristic polynomial and find the roots.
3. Make the general solution.
4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)
5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.
6. Substitute back into your general solution to acquire answer.
I am actually not 100% sure. My gut says it will be accepted, but I will need to speak to a senior marker to verify this.
That's epicPS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).
Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.
I'll summarise my method for the previous question in case I was ambiguous:
1. Make the characteristic polynomial.
2. Solve the characteristic polynomial and find the roots.
3. Make the general solution.
4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)
5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.
6. Substitute back into your general solution to acquire answer.
. I just did the Fibonnaci sequence with it. Thanks.
Thank you for that.
That's so much better!PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).
Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.
I'll summarise my method for the previous question in case I was ambiguous:
1. Make the characteristic polynomial.
2. Solve the characteristic polynomial and find the roots.
3. Make the general solution.
4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)
5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.
6. Substitute back into your general solution to acquire answer.
PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).
Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.
I'll summarise my method for the previous question in case I was ambiguous:
1. Make the characteristic polynomial.
2. Solve the characteristic polynomial and find the roots.
3. Make the general solution.
4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)
5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.
6. Substitute back into your general solution to acquire answer.
Like gurmies said, it is like DEs. Recursive relationships can be homogenous or inhomogenous etc. The method of solving them is usually taught in discrete maths, despite the method of auxiliary eqns.being so similar to that of integral calc for DEs.
I wanna see,
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $Prove true for n=1,2 assume true for $a_n$ and $a_{n-1}$ try to prove for $a_{n@plus;1}$ (Can't be bothered doing a proper proof at the moment)$\\ ~\\ \begin{align*} a_{n@plus;1}&=a_n@plus;6a_{n-1}\\ &=3^n-(-2)^n@plus;6(3^{n-1}-(-2)^{n-1})\\ &=3^{n-1}(3@plus;6)-(-2)^{n-1}(-2@plus;6)\\ &=3^{n-1}\cdot 3^2-(-2)^{n-1}\cdot (-2)^2\\ &=3^{n@plus;1}-(-2)^{n@plus;1} \end{align*}\\ $By the principle of mathematical induction blah blah blah it's true.$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $Prove true for n=1,2 assume true for $a_n$ and $a_{n-1}$ try to prove for $a_{n+1}$ (Can't be bothered doing a proper proof at the moment)$\\ ~\\ \begin{align*} a_{n+1}&=a_n+6a_{n-1}\\ &=3^n-(-2)^n+6(3^{n-1}-(-2)^{n-1})\\ &=3^{n-1}(3+6)-(-2)^{n-1}(-2+6)\\ &=3^{n-1}\cdot 3^2-(-2)^{n-1}\cdot (-2)^2\\ &=3^{n+1}-(-2)^{n+1} \end{align*}\\ $By the principle of mathematical induction blah blah blah it's true.$" title="\\ $Prove true for n=1,2 assume true for $a_n$ and $a_{n-1}$ try to prove for $a_{n+1}$ (Can't be bothered doing a proper proof at the moment)$\\ ~\\ \begin{align*} a_{n+1}&=a_n+6a_{n-1}\\ &=3^n-(-2)^n+6(3^{n-1}-(-2)^{n-1})\\ &=3^{n-1}(3+6)-(-2)^{n-1}(-2+6)\\ &=3^{n-1}\cdot 3^2-(-2)^{n-1}\cdot (-2)^2\\ &=3^{n+1}-(-2)^{n+1} \end{align*}\\ $By the principle of mathematical induction blah blah blah it's true.$" /></a>fun question
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