1. **Problem Statement:** Show that the sequence $\{a_n\}$ is a solution of the recurrence relation $$a_n = -3a_{n-1} + 4a_{n-2}$$ for the given sequences: 2. **Recurrence Relation:** The relation states that each term $a_n$ depends on the two previous terms $a_{n-1}$ and $a_{n-2}$ by the formula: $$a_n = -3a_{n-1} + 4a_{n-2}$$ 3. **Check each sequence:** **a) $a_n = 0$** - Substitute into the recurrence: $$0 = -3 \cdot 0 + 4 \cdot 0 = 0$$ - Holds true for all $n$. **b) $a_n = 1$** - Substitute: $$1 \stackrel{?}{=} -3 \cdot 1 + 4 \cdot 1 = -3 + 4 = 1$$ - Holds true for all $n$. **c) $a_n = (-4)^n$** - Substitute: $$(-4)^n \stackrel{?}{=} -3(-4)^{n-1} + 4(-4)^{n-2}$$ - Factor right side: $$= (-4)^{n-2}(-3(-4) + 4) = (-4)^{n-2}(12 + 4) = 16(-4)^{n-2}$$ - Left side: $$(-4)^n = (-4)^2 \cdot (-4)^{n-2} = 16(-4)^{n-2}$$ - Both sides equal, so true. **d) $a_n = 2(-4)^n + 3$** - Substitute: $$2(-4)^n + 3 \stackrel{?}{=} -3[2(-4)^{n-1} + 3] + 4[2(-4)^{n-2} + 3]$$ - Expand right side: $$= -6(-4)^{n-1} - 9 + 8(-4)^{n-2} + 12 = -6(-4)^{n-1} + 8(-4)^{n-2} + 3$$ - Factor powers: $$-6(-4)^{n-1} + 8(-4)^{n-2} = (-4)^{n-2}(-6(-4) + 8) = (-4)^{n-2}(24 + 8) = 32(-4)^{n-2}$$ - Left side: $$2(-4)^n + 3 = 2 \cdot 16 (-4)^{n-2} + 3 = 32(-4)^{n-2} + 3$$ - Both sides equal, so true. **Final conclusion:** All given sequences satisfy the recurrence relation.