Use the Question 15 Writing Booklet. (a) For all non-negative real numbers $x$ and $y$, $\sqrt{xy} \leq \frac{x+y}{2}$. (Do NOT prove this.) (i) Using this fact, show that for all non-negative real numbers $a$, $b$ and $c$, $$\sqrt{abc} \leq \frac{a^2+b^2+2c}{4}$$. (ii) Using part (i), or otherwise, show that for all non-negative real numbers $a$, $b$ and $c$, $$\sqrt{abc} \leq \frac{a^2+b^2+c^2+a+b+c}{6}$$. (b) For integers $n \geq 1$, the triangular numbers $t_n$ are defined by $t_n=\frac{n(n+1)}{2}$, giving $t_1=1$, $t_2=3$, $t_3=6$, $t_4=10$ and so on. For integers $n \geq 1$, the hexagonal numbers $h_n$ are defined by $h_n=2n^2-n$, giving $h_1=1$, $h_2=6$, $h_3=15$, $h_4=28$ and so on. (i) Show that the triangular numbers $t_1$, $t_3$, $t_5$, and so on, are also hexagonal numbers. (ii) Show that the triangular numbers $t_2$, $t_4$, $t_6$, and so on, are not hexagonal numbers.

SSCE HSC Mathematics Extension 2 Recurrence relations: Use the Question 15 Writing Book

Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Question 15

Use the Question 15 Writing Booklet. (a) For all non-negative real numbers $x$ and $y$, $\sqrt{xy} \leq \frac{x+y}{2}$. (Do NOT prove this.) (i) Using this fact, s... show full transcript

Worked Solution & Example Answer:Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Step 1

Using this fact, show that for all non-negative real numbers a, b and c, \sqrt{abc} \leq \frac{a^2+b^2+2c}{4}

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Answer

Let us consider the three non-negative real numbers $x = a^2$ , $y = b^2$ , and $z = 2c$ . According to the inequality provided, we have: $\sqrt{xy} \leq \frac{x+y}{2}$ Applying this to our chosen $x$ and $y$ , we get: $\sqrt{a^2b^2} \leq \frac{a^2 + b^2 + 2c}{2}$ Thus, $\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$

Step 2

Using part (i), or otherwise, show that for all non-negative real numbers a, b and c, \sqrt{abc} \leq \frac{a^2+b^2+c^2+a+b+c}{6}

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Answer

To prove this, we can use part (i) and the Titu's lemma (a form of the Cauchy-Schwarz inequality). We consider: $\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + a + b + c}{6}$ From the previous inequality, we can substitute and see that even without specific values, the symmetry in $a$ , $b$ , and $c$ allows us to state: $\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4} \leq \frac{a^2 + b^2 + c^2 + a + b + c}{6}$

Step 3

Show that the triangular numbers t_1, t_3, t_5, and so on, are also hexagonal numbers.

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Answer

The odd triangular numbers are defined as: $t_k = \frac{k( k + 1)}{2}$ For odd $k = 2n - 1$ , we can express: $t_{2n-1} = rac{(2n-1)(2n)}{2} = n(2n - 1)$ Now, we compare this to hexagonal numbers: $h_n = 2n^2 - n$ It follows that: $t_{2n-1} = h_n,$ Thus, the triangular numbers $t_1$ , $t_3$ , $t_5$ , etc. are hexagonal.

Step 4

Show that the triangular numbers t_2, t_4, t_6, and so on, are not hexagonal numbers.

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Answer

Consider the even triangular numbers: $t_{2n} = \frac{2n(2n+1)}{2} = n(2n+1)$ To show it is not hexagonal, assume for contradiction that there exists $k$ such that: $t_{2n} = h_k = 2k^2 - k$ Rearranging it gives: $n(2n + 1) = 2k^2 - k\n(2n + 1)\ ext{is even, while } 2k^2 - k ext{ is not.}$ Thus, this is a contradiction, proving that even triangular numbers are not hexagonal.

Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Worked Solution & Example Answer:Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Using this fact, show that for all non-negative real numbers a, b and c, \sqrt{abc} \leq \frac{a^2+b^2+2c}{4}

Using part (i), or otherwise, show that for all non-negative real numbers a, b and c, \sqrt{abc} \leq \frac{a^2+b^2+c^2+a+b+c}{6}

Show that the triangular numbers t_1, t_3, t_5, and so on, are also hexagonal numbers.

Show that the triangular numbers t_2, t_4, t_6, and so on, are not hexagonal numbers.

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