I do not know how to solve for Induction Conclusion A sequence has $x_1=8$, $x_2=32$, $x_{n}= 2x_{n-1}+3x_{n-2}$ for $n \ge 3.$ Prove, for all $i$ of Naturals, $X_i = 2 (-1)^i + 10 \cdot 3^{i-1}$
I got bases covered, and I got the inductive step as $X_{i} = 2 * (-1)^k + 10 * 3^{k-1}$. I do not know how to follow up with k+1, as I get complete gibberish. Any suggestions
Could someone guide me through each step?
My work so far:—
Base Cases include $x = 1, 2.$
$X_{1} = 2(-1)^1 + 10 \cdot 3^0$
$8 = 8$
$X_2 = 2(-1)^2 + 10 \cdot 3^{1}$
$32 = 32$
Inductive Step: $i = k$
$X_k = 2(-1)^k + 10 \cdot 3^{k-1}$
Inductive Conclusion:
$X_{k+1} = X_k + X_{k-1}$
$X_{k+1} = 2(2(-1)^k + 10 \cdot 3^{k-1}) + 3(2(-1)^{k-1} + 10 \cdot 3^{k-2})$
$X_{k+1} =  4(-1)^k + 6(-1)^{k-1} + 20 \cdot 3^{k-1} + 10 \cdot 3^{k-1}$
At this point. I know I should factor, but the 2nd half of $3^{k-1}$ does not create $10\cdot3^{k+1}$ as needed.
 A: You were close.
The base cases of $X_1 = 8$ and $X_2 = 32$ check out.
The formula to be verified is
$$X_k = [2 \times (-1)^{k}] + [10 \times 3^{(k-1)}].$$
The algorithm for expressing $X_{(k+1)}$ in terms of $X_k$ and $X_{(k-1)}$ is
$$X_{(k+1)} = [2 \times X_k] + [3 \times X_{(k-1)}].$$
Inductively assume that the formula holds for $X_k$ and $X_{(k-1)}.$
Then $$X_{(k+1)} = [2 \times X_k] + [3 \times X_{(k-1)}]$$
$$= 2 \times \{[2 \times (-1)^{k}] + [10 \times 3^{(k-1)}]\} + 
3 \times \{[2 \times (-1)^{(k-1)}] + [10 \times 3^{(k-2)}]\}$$
$$= [2 \times (-1)^{(k-1)} \times (3-2)] + [10 \times 3^{(k-2)} \times (3 + <2\times 3>)]$$
$$= [2 \times (-1)^{(k+1)}] + [10 \times 3^{(k-2)} \times (9)]$$
$$= [2 \times (-1)^{k+1)}] + [10 \times 3^{(k)}].$$
This completes the inductive step.
A: Let $A_n = x_n-x_{n-2}$ and $B_n = x_n+x_{n-1}$
We're given that $$x_n=2x_{n-1}+3x_{n-2}$$
$$\implies x_n-x_{n-2}=2(x_{n-1}+x_{n-2})$$
$$\implies A_n=2B_{n-1}$$
Also, using the definitions of $A_n$ and $B_n$:
$$B_n - A_n = B_{n-1}$$
Thus,
$$B_n = 3B_{n-1}$$
So, for general $B_n$:
$$B_n = 3^{n-2}\cdot B_{2}$$
$$\implies B_n = 3^{n-2}\cdot 40$$
Now we can use $B_n$ to calculate $x_n$.
$$\implies x_n+x_{n-1} = 3^{n-2}\cdot 40$$
And, $$ x_{n+1}+x_{n} = 3^{n-1}\cdot 40$$
So, subtracting the above equations:
$$x_{n+1}-x_{n-1} = 80 \cdot 3^{n-2}$$
And so,
$$x_{n+1} = x_1 + 80 \cdot (1+3^2+3^4+...+3^{n-2})$$
This is true if $n$ is even.
If $n$ is odd,
$$x_{n+1} = x_2 + 240 \cdot (1+3^2+...+3^{n-3})$$
Thus, for odd n (summing up the geometric progression):
$$x_{n+1}=10 \cdot 3^n +2$$
And for even n:
$$x_{n+1}=10 \cdot 3^n-2$$
Which implies that $x_n=10 \cdot 3^{n-1} + 2 \cdot (-1)^n$.
