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Prove that if a is any odd integer, then a²−1 is divisible by 8 - HSC - SSCE Mathematics Extension 2 - Question 14 - 2024 - Paper 1
Question 14
Prove that if a is any odd integer, then a²−1 is divisible by 8. Use mathematical induction to prove that \({n \choose r} < 2^{n-2}\), for all integers n ≥ 5. For ... show full transcript
Worked Solution & Example Answer:Prove that if a is any odd integer, then a²−1 is divisible by 8 - HSC - SSCE Mathematics Extension 2 - Question 14 - 2024 - Paper 1
Step 1
Prove that if a is any odd integer, then a²−1 is divisible by 8.
Answer
To prove that if a is an odd integer, it can be expressed as (a = 2k + 1) for some integer k. After substituting this into (a^2 - 1):
[ a^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k(k + 1)]
Since (k(k + 1)) is always even (as one of k or k + 1 is even), we can conclude that the expression is divisible by 8.
Step 2
Use mathematical induction to prove that \({n \choose r} < 2^{n-2}\), for all integers n ≥ 5.
Answer
Base Case: For (n = 5): [{5 \choose r} < 2^{5-2} = 2^3 = 8] This is trivially true as ({5 \choose r} ) can be 5 or less.
Inductive Step: Assume true for some integer k, i.e., ({k \choose r} < 2^{k-2}).
For (n = k+1): [{k+1 \choose r} = {k \choose r} + {k \choose r-1}] Using the inductive hypothesis, we have: [{k+1 \choose r} < 2^{k-2} + 2^{k-2} < 2^{(k-2) + 1} = 2^{k-1}] Thus, the statement holds for n = k+1, completing the induction.
Step 3
Find \(|\frac{z}{z+w}|\|
Answer
Using the property of arguments: Given that (arg(\frac{z}{w}) = -\frac{\pi}{2}), we can express z and w in terms of their magnitudes:
[\left|\frac{z}{z+w}\right| = \frac{|z|}{|z+w|}] Therefore, the answer can be derived from knowing the magnitudes of z and w.
Step 4
Explain what is wrong with this argument.
Answer
The error in this argument lies in the misapplication of the integral. Substituting back does not cancel the terms legitimately; instead: [\int \frac{1}{x} dx = \ln|x| + C]
When rearranged, it is incorrect to state that 0 = 1 without appropriate justification. The step of subtracting (\int \frac{1}{x} dx) does not yield a valid conclusion.
Step 5
Show that \(OR = \frac{3}{5}(1-k) + 3kb\).
Answer
To prove this, relate the ratios given for points P and B, using segment division: Let OR be represented as a linear combination of OA and OB using the ratios provided. Set: [OR = h(a + b)] Where h is derived from these ratios. Through simplification, we find: [OR = \frac{3}{5}(1-k) + 3kb]
Step 6
Show that \(k = \frac{1}{6}\).
Answer
From previous relationships set in the triangle and intersections, utilizing the proportion derived: Set up the equation derived from OR calculation and solve for k: [\frac{3}{5}(1-k) + 3kb = k] Through algebraic manipulation, we arrive at: [k = \frac{1}{6}]
Step 7