Final answer:
To prove the statement (a), we check the cases where 0 ≤ n < 3. For (n + 1)² > n³, we find that the statement holds true for n = 0, 1, and 2.
Step-by-step explanation:
To prove the statement (a), we can start by considering the values of 0, 1, and 2 for n.
For n = 0, we have (0 + 1)² = 1² = 1 and n³ = 0³ = 0. Since 1 > 0, the statement holds true for n = 0.
For n = 1, we have (1 + 1)² = 2² = 4 and n³ = 1³ = 1. Since 4 > 1, the statement holds true for n = 1.
For n = 2, we have (2 + 1)² = 3² = 9 and n³ = 2³ = 8. Since 9 > 8, the statement holds true for n = 2.
Therefore, for every integer n such that 0 ≤ n < 3, (n + 1)² > n³.
To prove the statement (b), we can similarly evaluate the values of 0, 1, 2, 3, and 4 for n.
For n = 0, we have 2⁽⁰⁺²⁾ = 2² = 4 and 3⁰ = 1. Since 4 > 1, the statement holds true for n = 0.
For n = 1, we have 2⁽¹⁺²⁾ = 2³ = 8 and 3¹ = 3. Since 8 > 3, the statement holds true for n = 1.
For n = 2, we have 2⁽²⁺²⁾ = 2⁴ = 16 and 3² = 9. Since 16 > 9, the statement holds true for n = 2.
For n = 3, we have 2⁽³⁺²⁾ = 2⁵ = 32 and 3³ = 27. Since 32 > 27, the statement holds true for n = 3.
For n = 4, we have 2⁽⁴⁺²⁾ = 2⁶ = 64 and 3⁴ = 81. Since 64 < 81, the statement does not hold true for n = 4.
Therefore, for every integer n such that 0 ≤ n ≤ 4, 2⁽ⁿ⁺²⁾ > 3ⁿ except for n = 4.