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Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + … + 2(−3)^{n−1} = − rac{1 − (−3)^{n}}{2} - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1
Question 13
Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + … + 2(−3)^{n−1} = − rac{1 − (−3)^{n}}{2}. The diagram shows the graph y = − rac{x}{x^{2} + 1}... show full transcript
Worked Solution & Example Answer:Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + … + 2(−3)^{n−1} = − rac{1 − (−3)^{n}}{2} - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1
Step 1
Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + … + 2(−3)^{n−1} = −\frac{1 − (−3)^{n}}{2}.
Answer
To prove the statement by induction, we proceed as follows:
Base Case (n=1): For n=1:
LHS = 2
RHS = −\frac{1 − (−3)^{1}}{2} = −\frac{1 + 3}{2} = 2
Thus, LHS = RHS, so the base case holds.
Induction Hypothesis: Assume the statement holds for n = k, i.e., 2 − 6 + 18 − 54 + ... + 2(−3)^{k−1} = −\frac{1 − (−3)^{k}}{2}.
Induction Step: We need to show that it holds for n = k + 1:
LHS = 2 − 6 + 18 − 54 + ... + 2(−3)^{k−1} + 2(−3)^{k} = −\frac{1 − (−3)^{k}}{2} + 2(−3)^{k}.
Combining terms yields:
LHS = −\frac{1 − (−3)^{k}}{2} + 2(−3)^{k} = −\frac{1 − (−3)^{k}}{2} + \frac{4(−3)^{k}}{2} = −\frac{1 + 3^{k + 1}}{2}.
RHS = −\frac{1 − (−3)^{k+1}}{2}.
Since LHS = RHS, this completes the proof by induction.
Step 2
State the domain and range of f^{−1}(x).
Answer
Domain of f^{−1}(x):
The domain of f^{−1}(x) corresponds to the range of f(x), which is all real numbers such that 0 < y < rac{1}{2}.
Range of f^{−1}(x):
The range of f^{−1}(x) corresponds to the domain of f(x), specifically all real numbers x ≥ 1.
Step 3
Sketch the graph y = f^{−1}(x).
Answer
To sketch the graph of y = f^{−1}(x):
- Mark critical points based on the information obtained from the domain and range.
- The graph will reflect the inverse relationship, hence it will be symmetric with respect to the line y = x.
- Begin the sketch at the point (1, − rac{1}{2}) and maintain the overall shape as guided by the properties of the original function.
Step 4
Find an expression for f^{−1}(x).
Answer
To find the expression for f^{−1}(x), start with:
-
Set y = f(x) = −\frac{x}{x^{2} + 1}.
-
Solve for x in terms of y:
Multiply both sides by (x^{2} + 1) to obtain:
(y(x^{2} + 1) = -x)
Rearranging gives:
(yx^{2} + x + y = 0)
Solve this quadratic equation for x using the quadratic formula:
(x = \frac{-1 \pm \sqrt{1 - 4y^{2}}}{2y})
Choose the appropriate sign to ensure x ≥ 1. This gives a valid expression for f^{−1}(x).