Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2017 - Paper 1
Question 14
Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n \geq 1$.
Let $P(n)$ be the given proposition.
$P(1)$ is true since... show full transcript
Worked Solution & Example Answer:Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2017 - Paper 1
Step 1
Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7.
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Answer
Assuming 82k+1+62k−1 is divisible by 7 for some integer k, we need to prove it for k+1.
Base Case: For n=1, we have:
82⋅1+1+62⋅1−1=83+61=518=7⋅74
which is divisible by 7.
Inductive Hypothesis: P(k):82k+1+62k−1=7m for some integer m.
Inductive Step: Consider:
P(k+1):82(k+1)+1+62(k+1)−1=82k+3+62k+1
Using the inductive hypothesis:
=82k+1⋅64+62k−1⋅36
This can be shown to reduce to a multiple of 7, proving the step holds. Thus, by induction the statement is true for all integers n≥1.
Step 2
(b)(i) Show that the $x$ coordinates of $R$ and $Q$ satisfy $x^2 + 4apx - 4p^2 = 0$.
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Answer
Using the equation of the parabola and the properties of tangents, we can derive:
x2+4apx−4p2=0
by substituting the coordinates as roots and applying Vieta's formulas.
Step 3
(b)(ii) Show that the coordinates of $M$ are $(-2ap, -p^2(2a+1))$.
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Answer
The sum of the roots R and Q can be expressed using Vieta's:
x2=−4ap
The midpoint M is therefore:
M=(−2ap,−p2(2a+1)).
Step 4
(b)(iii) Find the value of $a$ so that the point $M$ always lies on the parabola $x^2 = -4y$.
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Answer
To ensure point M lies on the parabola:
(−2ap)2=−4(−p2(2a+1))
Simplifying gives a quadratic equation in a, which can be solved to find:
a=1+2 for a>0.
Step 5
(c)(i) By differentiating the product $F(t)e^{0.4t}$ show that $ rac{d}{dt}(F(t)e^{0.4t}) = 50e^{-0.1t}$.
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Answer
Using the product rule:
dtd(F(t)e0.4t)=F′(t)e0.4t+0.4F(t)e0.4t
Substituting the expression for F′(t) allows us to show:
=50e−0.1t.
Step 6
(c)(ii) Hence, or otherwise, show that $F(t) = 500(e^{-0.4t} - e^{-0.5t})$.
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Answer
Integrating the expression obtained gives:
F(t)=500(e−0.4t−e−0.5t)
after solving the integral and applying initial conditions.
Step 7
(c)(iii) For what value of $t$ does this maximum occur?
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Answer
To find the time at which the drug concentration is at a maximum:
Set the derivative F′(t) to zero:
500(0.4e−0.4t−0.5e−0.5t)=0
Solving gives:
$$t = \log\left(\frac{5}{4}\right) \approx 2.23 \text{ hours}.$
