Let f(n)=A(-2)^n+B(-3)^n for all A, B in math | vedclass

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(D) Given $f(n) = A(-2)^n + B(-3)^n$. Substituting $f(n)$ into the equation $f(n) + a f(n-1) + b f(n-2) = 0$: $A(-2)^n + B(-3)^n + a(A(-2)^{n-1} + B(-

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$11$

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(D) Given $f(n) = A(-2)^n + B(-3)^n$. Substituting $f(n)$ into the equation $f(n) + a f(n-1) + b f(n-2) = 0$: $A(-2)^n + B(-3)^n + a(A(-2)^{n-1} + B(-3)^{n-1}) + b(A(-2)^{n-2} + B(-3)^{n-2}) = 0$. Grouping terms with $A$ and $B$: $A(-2)^{n-2} [(-2)^2 + a(-2) + b] + B(-3)^{n-2} [(-3)^2 + a(-3) + b] = 0$. $A(-2)^{n-2} [4 - 2a + b] + B(-3)^{n-2} [9 - 3a + b] = 0$. Since this holds for all $A, B$, the coefficients must be zero: $4 - 2a + b = 0 \implies b - 2a = -4$. $9 - 3a + b = 0 \implies b - 3a = -9$. Subtracting the two equations: $(b - 2a) - (b - 3a) = -4 - (-9) \implies a = 5$. Substituting $a=5$ into $b - 2a = -4$: $b - 10 = -4 \implies b = 6$. Finally, $(a+b)(b-a) = (5+6)(6-5) = 11 	imes 1 = 11$.

Let $f(n)=A(-2)^n+B(-3)^n$ for all $A, B \in \mathbb{R}$ and $n \in \mathbb{N}-\{1, 2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

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