IBDP Maths AHL 1.15 Proof by mathematical induction AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Markscheme
Base case: For \( n = 1 \), \( 5^2 – 2^3 = 25 – 8 = 17 \), which is divisible by 17.
Inductive step: Assume \( 5^{2k} – 2^{3k} \) is divisible by 17.
For \( n = k + 1 \):
\( 5^{2(k+1)} – 2^{3(k+1)} = 25 \times 5^{2k} – 8 \times 2^{3k} \)
This can be written as \( 25(5^{2k} – 2^{3k}) + 17 \times 2^{3k} \), which is divisible by 17.
Therefore, by induction, \( 5^{2n} – 2^{3n} \) is divisible by 17 for all \( n \in \mathbb{Z}^+ \).
▶️ Answer/Explanation
Solution
Assume that \( a \) and \( b \) are both odd.
Let \( a = 2m + 1 \) and \( b = 2n + 1 \) where \( m, n \in \mathbb{Z} \).
Then:
\[ a^2 + b^2 = (2m + 1)^2 + (2n + 1)^2 \] \[ = 4m^2 + 4m + 1 + 4n^2 + 4n + 1 \] \[ = 4(m^2 + m + n^2 + n) + 2 \]Since \( 4(m^2 + m + n^2 + n) \) is divisible by 4, but 2 is not divisible by 4:
\[ a^2 + b^2 \equiv 2 \pmod{4} \]This contradicts the given condition that \( a^2 + b^2 \) is divisible by 4.
Therefore, our assumption must be false, and \( a \) and \( b \) cannot both be odd.
▶️ Answer/Explanation
Solution (METHOD 1 – Rearranging the equation)
Assume there exists some \( \alpha \in \mathbb{Z} \) such that:
\[ 2\alpha^3 + 6\alpha + 1 = 0 \]Rearranging the equation:
\[ 2\alpha^3 + 6\alpha = -1 \] \[ 2\alpha(\alpha^2 + 3) = -1 \]Since \( \alpha \) is integer, \( \alpha^2 + 3 \) is also integer. Therefore:
\[ \alpha \text{ must divide } -1 \text{ (i.e., } \alpha = \pm 1) \]Testing possible integer values:
For \( \alpha = 1 \):
\[ 2(1)^3 + 6(1) + 1 = 2 + 6 + 1 = 9 \neq 0 \]For \( \alpha = -1 \):
\[ 2(-1)^3 + 6(-1) + 1 = -2 – 6 + 1 = -7 \neq 0 \]No integer values satisfy the original equation, leading to a contradiction.
Therefore, the equation \( 2x^3 + 6x + 1 = 0 \) has no integer roots.
Markscheme Notes
Note: Award M1 for assumption like “assume \( \alpha \) is an integer root of \( 2\alpha^3 + 6\alpha + 1 = 0 \)”.
Note: Award M0 for vague statements like “let’s consider the equation has integer roots…”.
Note: Subsequent marks after the initial assumption are independent and can be awarded.
▶️ Answer/Explanation
Solution
Let \( f(n) = 5^n + 9^n + 2 \) and let \( P_n \) be the proposition that \( f(n) \) is divisible by 4.
Base Case (n=1):
\[ f(1) = 5^1 + 9^1 + 2 = 5 + 9 + 2 = 16 \]Since 16 is divisible by 4, \( P_1 \) is true.
Inductive Step:
Assume \( P_k \) is true for some \( k \in \mathbb{Z}^+ \), i.e., \( f(k) \) is divisible by 4.
Consider \( f(k+1) \):
\[ f(k+1) = 5^{k+1} + 9^{k+1} + 2 \] \[ = 5 \cdot 5^k + 9 \cdot 9^k + 2 \] \[ = (4+1)5^k + (8+1)9^k + 2 \] \[ = 4 \cdot 5^k + 8 \cdot 9^k + (5^k + 9^k + 2) \] \[ = 4(5^k + 2 \cdot 9^k) + f(k) \]Both terms are divisible by 4:
- \( 4(5^k + 2 \cdot 9^k) \) is clearly divisible by 4
- \( f(k) \) is divisible by 4 by the induction hypothesis
Therefore, \( f(k+1) \) is divisible by 4.