Given {U_{n + 1}} = 3{U_n} - 2{U_{n - 1}} and | vedclass

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(A) Step $I$: Given the recurrence relation ${U_{n + 1}} = 3{U_n} - 2{U_{n - 1}}$ with ${U_0} = 2$ and ${U_1} = 3$.For $n = 1$,${U_2} = 3{U_1} - 2{U_0

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${2^n} + 1$

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(A) Step $I$: Given the recurrence relation ${U_{n + 1}} = 3{U_n} - 2{U_{n - 1}}$ with ${U_0} = 2$ and ${U_1} = 3$.For $n = 1$,${U_2} = 3{U_1} - 2{U_0} = 3(3) - 2(2) = 9 - 4 = 5$.Checking option $(A)$ ${U_n} = {2^n} + 1$:For $n = 0$,${U_0} = {2^0} + 1 = 1 + 1 = 2$ (True).For $n = 1$,${U_1} = {2^1} + 1 = 3$ (True).For $n = 2$,${U_2} = {2^2} + 1 = 5$ (True).Step $II$: Assume ${U_k} = {2^k} + 1$ and ${U_{k - 1}} = {2^{k - 1}} + 1$.Step $III$: Using the recurrence relation for $n = k$:${U_{k + 1}} = 3{U_k} - 2{U_{k - 1}}$${U_{k + 1}} = 3({2^k} + 1) - 2({2^{k - 1}} + 1)$${U_{k + 1}} = 3 \cdot {2^k} + 3 - 2 \cdot {2^{k - 1}} - 2$${U_{k + 1}} = 3 \cdot {2^k} - {2^k} + 1$${U_{k + 1}} = (3 - 1) \cdot {2^k} + 1 = 2 \cdot {2^k} + 1 = {2^{k + 1}} + 1$.Thus,by the principle of mathematical induction,${U_n} = {2^n} + 1$ for all $n \in N \cup \{0\}$.

Given ${U_{n + 1}} = 3{U_n} - 2{U_{n - 1}}$ and ${U_0} = 2$,${U_1} = 3$,the value of ${U_n}$ for all $n \in N$ is

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