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. 1. For $n = 1$, $P(1)$ is true since $8^3 + 6^1 = 512 + 6 = 518$, which is divisible by 7. 2. Assume $P(k)$ is true for some integer $k$. That is, $8^{2k+1} + 6^{2k-1}$ is divisible by 7. This means there exists an integer $m$ such that: $$8^{2k+1} + 6^{2k-1} = 7m.$$ 3. Now we need to show $P(k+1)$: $$8^{2(k+1)+1} + 6^{2(k+1)-1} = 8^{2k+3} + 6^{2k+1}$$ $$= 8^{2k+1} \cdot 8^2 + 6^{2k-1} \cdot 6^2$$ $$= 64 \cdot 8^{2k+1} + 36 \cdot 6^{2k-1}$$ $$= 64(7m - 6^{2k-1}) + 36 \cdot 6^{2k-1}$$ $$= 448m - 384 + 36 \cdot 6^{2k-1}$$ $$= 448m - 384 + 36m = 7(64m + 6^{2k-1})$$ 4. Since this expression is also of the form $7n$ for some integer $n$, $P(k+1)$ is true. Hence by induction, $P(n)$ holds for all integers $n \geq 1$.

SSCE HSC Mathematics Extension 1 Proving divisibility by induction: Prove by mathematical induction

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

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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. 1. For $n = 1$, $P(1... 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

Let P(2p, p^2) be a point on the parabola x^2 = 4y.

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The coordinates for the point are given as $P(2p, p^2)$ . The tangent at point $P$ meets the parabola $x^2 = -4ay$ and can be derived from the slope of the curve at point $P$ . This tangent line can be expressed as:

The equation of the tangent at $P$ can be inferred by substituting into the parabola's equation: $y - p^2 = \frac{1}{2p}(x - 2p)$ Simplifying this yields the equation of the tangent line.
To show that the x-coordinates of points $Q$ and $R$ satisfy the equation $x^2 + 4apx - 4p^2 = 0$ , we will substitute the derived equation of the tangent into the mentioned equation and verify that it holds true.

Step 2

Show that the coordinates of M are (-2ap, -p^2(2a + 1)).

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Using the properties of the roots, the coordinates of $M$ can be calculated as the average of the roots: $x_M = -2ap,\ y_M = -p^2(2a+1).$ Here we find that $M$ is positioned at the average of the endpoints Q and R, thus verifying the coordinates.

Step 3

Find the value of a so that the point M always lies on the parabola x^2 = -4ay.

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To ensure that point $M$ with coordinates $(-2ap, -p^2(2a + 1))$ lies on the parabola, we substitute these coordinates into the parabola equation: $(-2ap)^2 = -4a(-p^2(2a + 1)),$ leading us to derive a condition on $a$ : $4a^2p^2 + 8ap^2 + 4ap^2 = 0$ Simplifying this yields: $a = 1 + \sqrt{2}.$

Step 4

The concentration of a drug in a body is F(t), where t is the time in hours after the drug is taken.

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To determine the rate of change of the drug concentration: Initially, the concentration is zero. The rate is given by: $F'(t) = 50e^{-0.5t} - 0.4F(t).$ This expression can be differentiated and analyzed to find the maximum concentration of the drug over time using calculus.

Step 5

By differentiating the product F(t)e^{0.4t} show that \frac{d}{dt}[F(t)e^{0.4t}] = 50e^{-0.1t}.

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Applying the product rule: $\frac{d}{dt}[F(t)e^{0.4t}] = F'(t)e^{0.4t} + F(t)(0.4e^{0.4t}).$ Substituting the expression for $F'(t)$ , we can simplify to derive: $= e^{0.4t}(50e^{-0.5t}).$

Step 6

Hence, or otherwise, show that F(t) = 500(e^{-0.4t} - e^{-0.5t}).

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Integrating the expression gives: $F(t)e^{0.4t} = -500e^{-0.1t} + c$ Evaluating the integration constant to find: $F(t) = 500(e^{-0.4t} - e^{-0.5t}).$

Step 7

The concentration of the drug increases to a maximum.

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To find when this occurs, set: $F'(t) = 500(0.4e^{-0.4t} - 0.5e^{-0.5t}) = 0.$ Solving gives: $0.4e^{-0.4t} = 0.5e^{-0.5t}.$ Leading to: $t = \ln(\frac{5}{4})/0.1.$ This provides the corresponding time at which the concentration is maximized.

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

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

Let P(2p, p^2) be a point on the parabola x^2 = 4y.

Show that the coordinates of M are (-2ap, -p^2(2a + 1)).

Find the value of a so that the point M always lies on the parabola x^2 = -4ay.

The concentration of a drug in a body is F(t), where t is the time in hours after the drug is taken.

By differentiating the product F(t)e^{0.4t} show that \frac{d}{dt}[F(t)e^{0.4t}] = 50e^{-0.1t}.

Hence, or otherwise, show that F(t) = 500(e^{-0.4t} - e^{-0.5t}).

The concentration of the drug increases to a maximum.

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