# Math Help - a=3n^(n-1)+2(-1)^n for neN

1. ## a=3n^(n-1)+2(-1)^n for neN

I was wondering if anyone could help me with the following question:

Thanks,

2. Try induction, it's real easy in this case.

3. Originally Posted by Jose27
Try induction, it's real easy in this case.
ohh ok
So would i use n=3 as my basecase and the sub k+1 for all n ?

4. Originally Posted by jess0517
ohh ok
So would i use n=3 as my basecase and the sub k+1 for all n ?
Basically, just prove the case n=3, then assume it's valid for n=k and, using this, prove that it is valid for k+1.

5. Hello, jess0517!

We can prove it by deriving that function ourselves . . .

$\text{Let }\{a\} \text{ be a sequence satisfying:}$

. . $a_1 = 1,\;\;a_2 = 8,\;\;a_n \:=\:a_{n-1} + 2a_{n-2}\:\text{ for }n \ge 3.$

$\text{Prove that: }\;a_n \;=\;3\!\cdot\!2^{n-1} + 2(\text{-}1)^n\:\text{ for }n \in N.$

We have: . $a_n - a_{n-1} - 2a_{n-2} \;=\;0$

Let $X^n = a_n$

. . and we have: . $X^n - X^{n-1} - 2X^{n-2} \:=\:0$

Divide by $X^{n-2}\!:\;\;X^2 - X - 2 \:=\:0$

. . Then: . $(X - 2)(X+1) \:=\:0 \quad\Rightarrow\quad X \:=\:2,\:-1$

The generating function is of the form: . $f(n) \;=\;A\!\cdot\!2^n + B\!\cdot\!(\text{-}1)^n$

We know the first two values of the sequence:

. . $\begin{array}{cccccccccc}
f(1) = 1\!: && 2A - B &=& 1 & [1] \\
f(2) = 8\!: && 4A + B &=& 8 & [2] \end{array}$

Add [1] and [2]: . $6A \:=\:9 \quad\Rightarrow\quad A \:=\:\frac{3}{2}$

Substitute into [1]: . $2(\frac{3}{2}) - B \:=\:1 \quad\Rightarrow\quad B \:=\:2$

Hence: . $f(n) \;=\;\frac{3}{2}\!\cdot\!2^n + 2(\text{-}1)^n$

Therefore: . $a_n \;=\;3\!\cdot\!2^{n-1} + 2(\text{-}1)^n$

6. Thread closed due to this member deleting questions after getting help.