Anyone can give me hints on this (separate)
- Thread starter Mono Yung
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There is a "typo" in the problem statement. It should state:
".... such that a1 = 2, a2 =20, ......"
The first problem contains a mistake.
[MATH]a_1 = 2.[/MATH]
So validate for n = 1. It is a tricky little problem because a1 is not 1 so it is easy to get mixed up. Even the person who wrote (or proof read) the problem got mixed up. I think there are two ways to solve it. The obvious way is by induction. I suspect that may be the easier way.
[MATH]a_1 = 2.[/MATH]
So validate for n = 1. It is a tricky little problem because a1 is not 1 so it is easy to get mixed up. Even the person who wrote (or proof read) the problem got mixed up. I think there are two ways to solve it. The obvious way is by induction. I suspect that may be the easier way.
HallsofIvy
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Yes, given that the solution is \(\displaystyle a_n= 4^n+ (-2)^n\) you would use "induction" to prove that it satisfies the equation \(\displaystyle a_{n+2}= 2a_{n+1}+ 8a_n\). But frankly I think that is "overkill". Just write \(\displaystyle a_{n+1}= 4^{n+1}+ (-2)^{n+1}= 4(4^n)- 2(-2)^n\) and \(\displaystyle a_{n+2}= 4^{n+ 2}+ (-2)^{n+2}= 16(4^n)+ 4(-2)^n\) so that \(\displaystyle 2a_{n+1}+ 8a_n= 8(4^n)- 4(-2)^n+ 8(4^n)+ 8(-2)^n= 16(4^n)+ 4(-2)^n\).
If, however, you were given the equation and asked to solve it you might argue that, since this involves multiplying by integers you should "try" a solution of the form \(\displaystyle a_n= r^n\). Putting that into the equation, \(\displaystyle r^{n+2}= 2r^{n+1}+ 8r^n\). Divide by \(\displaystyle r^n\) to reduce to \(\displaystyle r^2= 2r+ 8\) so that \(\displaystyle r^2- 2r+ 1= 9\). Then \(\displaystyle (r- 1)^2= 9\) so \(\displaystyle r-1= \pm 3\). Taking "+3", r= 4. Taking -3, r= -2. That gives us the two solutions, \(\displaystyle 4^n\) and \(\displaystyle (-2)^n\). Since \(\displaystyle a_{n+2}= 2a_{n+1}+ 8a_n\) is a linear equation, we could actually say that any function \(\displaystyle a_n= A4^n+ B(-2)^n\) is a solution for any constants, A and B.[/tex]
If, however, you were given the equation and asked to solve it you might argue that, since this involves multiplying by integers you should "try" a solution of the form \(\displaystyle a_n= r^n\). Putting that into the equation, \(\displaystyle r^{n+2}= 2r^{n+1}+ 8r^n\). Divide by \(\displaystyle r^n\) to reduce to \(\displaystyle r^2= 2r+ 8\) so that \(\displaystyle r^2- 2r+ 1= 9\). Then \(\displaystyle (r- 1)^2= 9\) so \(\displaystyle r-1= \pm 3\). Taking "+3", r= 4. Taking -3, r= -2. That gives us the two solutions, \(\displaystyle 4^n\) and \(\displaystyle (-2)^n\). Since \(\displaystyle a_{n+2}= 2a_{n+1}+ 8a_n\) is a linear equation, we could actually say that any function \(\displaystyle a_n= A4^n+ B(-2)^n\) is a solution for any constants, A and B.[/tex]