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Let $f(n)=A(-2)^n+B(-3)^n \forall A, B \in \mathbf{R}$ and $n \in \mathbf{N}-\{1,2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$
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Verified Answer
The correct answer is:
11
$\begin{aligned}
& \text { We have, } f(n)=A(-2)^n+B(-3)^n \\
& \qquad \begin{array}{r}
f(n)+a f(n-1)+b f(n-2)=0 \\
\therefore \quad A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right) \\
\\
\quad+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right. \\
\Rightarrow A(-2)^{n-2}[4-2 a+b]+(-3)^{n-2} B[9-3 a+b]=0
\end{array}
\end{aligned}$
$\therefore$ It is possible only
$4-2 a+b=0 \text { and } 9-3 a+b=0$
Solving, we get $a=5, b=6$
$\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11$
& \text { We have, } f(n)=A(-2)^n+B(-3)^n \\
& \qquad \begin{array}{r}
f(n)+a f(n-1)+b f(n-2)=0 \\
\therefore \quad A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right) \\
\\
\quad+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right. \\
\Rightarrow A(-2)^{n-2}[4-2 a+b]+(-3)^{n-2} B[9-3 a+b]=0
\end{array}
\end{aligned}$
$\therefore$ It is possible only
$4-2 a+b=0 \text { and } 9-3 a+b=0$
Solving, we get $a=5, b=6$
$\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11$
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