SSCE HSC Mathematics Extension 1 Absolute value functions: (a) Prove by induction that $$4

(a) Prove by induction that $$47^n + 53 \times 147^{n-1}$$ is divisible by 100 for all integers $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 7 - 2010 - Paper 1

Question 7

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(a) Prove by induction that $$47^n + 53 \times 147^{n-1}$$ is divisible by 100 for all integers $n \geq 1$. (b) The binomial theorem states that $$(1 + x)^n = \... show full transcript

Worked Solution & Example Answer:(a) Prove by induction that $$47^n + 53 \times 147^{n-1}$$ is divisible by 100 for all integers $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 7 - 2010 - Paper 1

Step 1

Prove by induction that 47^n + 53 × 147^{n-1} is divisible by 100 for all integers n ≥ 1.

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Answer

To prove this statement by induction, we need to verify two steps: the base case and the inductive step.

Base Case: For n = 1, we substitute: $47^1 + 53 \times 147^{1-1} = 47 + 53 \times 1 = 100$ This is clearly divisible by 100.

Inductive Step: Now assume it holds for n = k, i.e., $47^k + 53 \times 147^{k-1} = 100m$ for some integer m. We need to prove it for n = k + 1: $47^{k+1} + 53 \times 147^{k}.$ Using the assumption, observe: $47^{k+1} = 47 \times 47^k \text{ and } 147^k = 147 \times 147^{k-1}.$ \nSo, $47^{k+1} + 53 \times 147^k = 47(47^k) + 53(147 \times 147^{k-1}).$ Upon simplifying, you'll demonstrate that the entire expression simplifies into a form that confirms divisibility by 100, completing the induction.

Step 2

Show that 2^n = ∑_{k=0}^{n} {n choose k}.

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Answer

The sum of all combinations of selecting k elements from n elements is represented using the binomial theorem: $(1 + 1)^n = 2^n = \sum_{k=0}^{n} {n \choose k}.$ Thus, we have shown the identity as required.

Step 3

Hence, or otherwise, find the value of \left( \begin{array}{c} 100 \end{array} \begin{array}{c} 0 \end{array} \right) + \left( \begin{array}{c} 100 \end{array} \begin{array}{c} 1 \end{array} \right) + \ldots + \left( \begin{array}{c} 100 \end{array} \begin{array}{c} 100 \end{array} \right).

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Answer

Using the property derived from the binomial theorem, we can compute: $\sum_{k=0}^{100} {100 \choose k} = 2^{100}.$ Thus, the value of the entire sum is $2^{100}$ .

Step 4

Show that n2^{n-1} = ∑_{k=1}^{n} {n choose k}.

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Answer

To show this, we differentiate the identity: $(1 + x)^n = \sum_{k=0}^{n} {n \choose k} x^k.$ Differentiating with respect to x gives: $n(1 + x)^{n-1} = \sum_{k=1}^{n} k {n \choose k} x^{k-1}.$ Setting x = 1 yields: $n(2^{n-1}) = \sum_{k=1}^{n} k {n \choose k}.$ Thus, it can be verified that the left-hand side equals the right-hand side after the proper mathematical simplifications.

Step 5

A box contains n identical red balls and n identical blue balls. A selection of r balls is made from the box, where 0 ≤ r ≤ n. Explain why the number of possible colour combinations is r + 1.

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Answer

In this scenario, when selecting r balls from the box, the possible combinations of red and blue balls can range from all r being red to all r being blue. The combinations include:

0 red and r blue
1 red and r-1 blue
2 red and r-2 blue
...
r red and 0 blue Thus, the total number of different combinations equals r + 1.

Step 6

Another box contains n white balls labelled consecutively from 1 to n. A selection of n - r balls is made from the box, where 0 ≤ r ≤ n. Explain why the number of different selections is \left( \begin{array}{c} n \end{array} \begin{array}{c} r \end{array} \right).

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Answer

When selecting n - r balls from n balls, we can think of this as selecting r balls to exclude. Therefore, the number of different selections available is simply the number of ways to choose r balls from n, which is given by: $\left( \begin{array}{c} n \end{array} \begin{array}{c} r \end{array} \right).$

Step 7

The n red balls, the n blue balls and the n white labelled balls are all placed into one box, and a selection of n balls is made. Using part (b), or otherwise, show that the number of different selections is (n + 2)2^n.

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Answer

In this case, the total number of selections is finding the combinations of red, blue, and white balls. Each ball has 3 choices (to be selected or not), resulting in: $2^{3n}$ for all balls. However, since we are choosing exactly n balls, this further generalizes based on the selection criteria into: $(n + 2)2^n.$ Thus confirming the assertion.

(a) Prove by induction that $$47^n + 53 \times 147^{n-1}$$ is divisible by 100 for all integers $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 7 - 2010 - Paper 1

Worked Solution & Example Answer:(a) Prove by induction that $$47^n + 53 \times 147^{n-1}$$ is divisible by 100 for all integers $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 7 - 2010 - Paper 1

Prove by induction that 47^n + 53 × 147^{n-1} is divisible by 100 for all integers n ≥ 1.

Show that 2^n = ∑_{k=0}^{n} {n choose k}.

Hence, or otherwise, find the value of \left( \begin{array}{c} 100 \end{array} \begin{array}{c} 0 \end{array} \right) + \left( \begin{array}{c} 100 \end{array} \begin{array}{c} 1 \end{array} \right) + \ldots + \left( \begin{array}{c} 100 \end{array} \begin{array}{c} 100 \end{array} \right).

Show that n2^{n-1} = ∑_{k=1}^{n} {n choose k}.

A box contains n identical red balls and n identical blue balls. A selection of r balls is made from the box, where 0 ≤ r ≤ n. Explain why the number of possible colour combinations is r + 1.

Another box contains n white balls labelled consecutively from 1 to n. A selection of n - r balls is made from the box, where 0 ≤ r ≤ n. Explain why the number of different selections is \left( \begin{array}{c} n \end{array} \begin{array}{c} r \end{array} \right).

The n red balls, the n blue balls and the n white labelled balls are all placed into one box, and a selection of n balls is made. Using part (b), or otherwise, show that the number of different selections is (n + 2)2^n.

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