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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 e... 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 the statement using induction, we first verify the base case (n=1):
21+(1)1+1=2+1=32^1 + (-1)^{1+1} = 2 + 1 = 3
which is divisible by 3.

Next, assume that the statement holds for some integer k1k \geq 1:
2k+(1)k+1 is divisible by 3.2^k + (-1)^{k+1} \text{ is divisible by } 3.
We need to show that it holds for k+1k + 1:
2k+1+(1)(k+1)+1=2×2k+(1)k+2=2×2k1.2^{k+1} + (-1)^{(k+1)+1} = 2 \times 2^k + (-1)^{k+2} = 2 \times 2^k - 1.
Adding the induction hypothesis:
2×2k1+(2k+(1)k+1)=(2+1)×2k1=3×2k1.2 \times 2^k - 1 + (2^k + (-1)^{k+1}) = (2 + 1) \times 2^k - 1 = 3 \times 2^k - 1.
We know 3×2k3 \times 2^k is divisible by 3, and 1+3×2k-1 + 3 \times 2^k ensures that the entire expression is divisible by 3, making the statement valid for k + 1. Thus, the induction step holds and the proof is complete.

Step 2

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

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Answer

Using the Pythagorean theorem, we have:
L2=x2+402.L^2 = x^2 + 40^2.
Differentiating both sides with respect to xx:
2LdLdx=2x,2L \frac{dL}{dx} = 2x,
which simplifies to:
dLdx=xL.\frac{dL}{dx} = \frac{x}{L}.
Since cosθ=xL\cos\theta = \frac{x}{L}, we can conclude that:
dLdx=cosθ.\frac{dL}{dx} = \cos\theta.

Step 3

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

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Answer

Applying the chain rule:
dLdt=dLdxdxdt.\frac{dL}{dt} = \frac{dL}{dx} \cdot \frac{dx}{dt}.
From the prior result, we have dLdx=cosθ\frac{dL}{dx} = \cos\theta and since the truck moves right at a speed of 3 m/s, then:
dxdt=3.\frac{dx}{dt} = 3.
Combining these results yields:
dLdt=cosθ3=3cosθ.\frac{dL}{dt} = \cos\theta \cdot 3 = 3 \cos\theta.

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