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Use mathematical induction to prove that $2^n + (-1)^{n+1}$ is divisible by 3 for all integers $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 13 - 2014 - Paper 1

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Use mathematical induction to prove that $2^n + (-1)^{n+1}$ is divisible by 3 for all integers $n \geq 1$. One end of a rope is attached to a truck and the other en... show full transcript

Worked Solution & Example Answer:Use mathematical induction to prove that $2^n + (-1)^{n+1}$ is divisible by 3 for all integers $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 13 - 2014 - Paper 1

Step 1

Use mathematical induction to prove that $2^n + (-1)^{n+1}$ is divisible by 3 for all integers $n \geq 1$

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Answer

To prove that 2n+(1)n+12^n + (-1)^{n+1} is divisible by 3 for all integers n1n \geq 1, we can employ mathematical induction.

Base Case: For n=1n = 1, we calculate: 21+(1)1+1=2+1=3,2^1 + (-1)^{1+1} = 2 + 1 = 3, which is clearly divisible by 3.

Inductive Step: Assume the statement holds for some integer kk, i.e., that 2k+(1)k+12^k + (-1)^{k+1} is divisible by 3. We must show it is also true for k+1k + 1: 2k+1+(1)(k+1)+1=22k+(1)k+2=2(2k)1.2^{k+1} + (-1)^{(k+1)+1} = 2 \cdot 2^k + (-1)^{k+2} = 2(2^k) - 1.

From the inductive hypothesis, we know that 2k+(1)k+1=3m2^k + (-1)^{k+1} = 3m for some integer mm. Therefore, substituting gives us: 2(2k+(1)k+1)1=2(3m)1=6m1.2(2^k + (-1)^{k+1}) - 1 = 2(3m) - 1 = 6m - 1.

We need to show that 6m16m - 1 is also divisible by 3. Note that: 6m1=3(2m)1,6m - 1 = 3(2m) - 1, which is not directly useful for divisibility by 3. To finalize: 2k+1+(1)k+2=2(3m)1=6m1=3(2m1).2^{k+1} + (-1)^{k+2} = 2(3m) - 1 = 6m - 1 = 3(2m-1). Thus, this expression is divisible by 3. Hence, by the principle of mathematical induction, the statement is proven for all integers n1n \geq 1.

Step 2

Using Pythagoras' Theorem, or otherwise, show that $\frac{dL}{dx} = \cos \theta$

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Answer

Consider the right triangle formed by the vertical distance (40 m), the horizontal distance (xx m), and the distance to the wheel (LL m). Using Pythagorean Theorem, we have: L2=x2+402.L^2 = x^2 + 40^2.

Differentiating both sides with respect to xx, we get: 2LdLdx=2x,2L \frac{dL}{dx} = 2x, which simplifies to: dLdx=xL.\frac{dL}{dx} = \frac{x}{L}.

Also, from the definition of cosine in the triangle, we have: cosθ=xL.\cos \theta = \frac{x}{L}.

Thus: $$\frac{dL}{dx} = \cos \theta.$

Step 3

Show that $\frac{dL}{dt} = 3 \cos \theta$

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Answer

Using the chain rule, we can relate dLdt\frac{dL}{dt} and dLdx\frac{dL}{dx}: dLdt=dLdxdxdt.\frac{dL}{dt} = \frac{dL}{dx} \cdot \frac{dx}{dt}.

From the previous part, we have: dLdx=cosθ,\frac{dL}{dx} = \cos \theta, and since the truck is moving to the right at a constant speed of 3 m/s, we know: dxdt=3.\frac{dx}{dt} = 3.

Substituting these into the equation gives us: dLdt=cosθ3=3cosθ.\frac{dL}{dt} = \cos \theta \cdot 3 = 3 \cos \theta.

Hence, this demonstrates the required relationship.

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