If $P_n=\alpha^n+\beta^n\;, \alpha+\beta=1, \;\alpha \cdot \beta=-1,\;P_{n-1}=11,\; P_{n+1}=29$, $\alpha$ and $\beta$ are real numbers. Find $(P_n)^2,\;$ where $n\in \mathbb N$
My Approach:
Method $1$ :
$P_{n-1}\cdot P_{n+1}=11\cdot 29 \implies\; (\alpha^{n-1}+\beta^{n-1})\cdot (\alpha^{n+1}+\beta^{n+1})=\alpha^{2n}+\beta^{2n}+(\alpha \beta)^{n-1}\cdot(\alpha^2+\beta^2)$
$\implies \; 319=P_{2n}+3 (-1)^{n-1}$
$\implies\;P_{2n}=319-3(-1)^{n-1} \implies P_{2n}=319+3(-1)^n$
Now
$(P_{n})^2=(\alpha^n+\beta^n)^2=\alpha^{2n}+\beta^{2n}+2(\alpha \beta)^n\;\implies\;(P_{n})^2=P_{2n}+2(-1)^n$
$\implies \; (P_{n})^2=319+5(-1)^n$
$\implies (P_n)^2=324\;$ If $n$ is even and $(P_n)^2=314\;$ if $n$ is odd.
But given answer is $324$ only.
Method $2$:
Form a quadratic equation with sum of roots and product of roots is given
$\alpha^2-\alpha-1=0\quad$ and $\quad\beta^2-\beta-1=0$
$\implies \;\alpha^{n+1}=\alpha^n+\alpha^{n-1}\cdots(1)\quad$ and $\quad \beta^{n+1}=\beta^{n}+\beta^{n-1}\cdots (2)$
Now add both equation $(1)+(2)$
$\implies$ $P_{n+1}=P_{n}+P_{n-1} \; \implies \; P_{n}=18$
$\implies \; (P_n)^2=324$.
My doubt: What is wrong with my method $1$ ?
Since $P_n = \alpha^n + \beta^n $, then $P_n$ satisfies the difference equation
$ (E - \alpha)(E - \beta) P_n = 0 $
where $E$ is the advance operator, i.e. $E\left(P_n\right) = P_{n+1}$
Hence, multiplying the operators, we get the difference equation
$P_{n+2} - (\alpha + \beta) P_{n+1} + \alpha \beta P_{n} = 0 $
Substituting the given values of $(\alpha + \beta)$ and $\alpha \beta$,
$P_{n+2} - P_{n+1} - P_{n} = 0 $
That is,
$P_{n+2} = P_{n+1} + P_n$
Now
$P_1 = \alpha + \beta = 1 $
$P_2 = \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2 \alpha \beta = 3 $
Therefore, the sequence $\{P_n\}$ is as follows:
$1, 3, 4, 7, 11, 18, 29, 47, ... $
We already see our terms $11$ and $29$ , hence the one between them is $ 18 $, whose square is $324$