Anyone can give me hints on this (separate)

[Mono Yung]

can anyone give me hints on


thanks.

 

[lev888]

Tried math induction?

 

[Deleted member 4993]

can anyone give me hints on

View attachment 23995
thanks.

There is a "typo" in the problem statement. It should state:

".... such that a₁ = 2, a₂ =20, ......"

 

[JeffM]

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.

 

[HallsofIvy]

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]