Use mathematical induction to prove that, for $n \geq 1$, $$x^{(3n)} - 1 = \left( x - 1 \right) \left( x^{2n} + x^{n+1} \right) \cdots \left( x^{(n-1)(2n+1)} + x^{(n-1)(3)} + 1 \right)$$ In $\triangle ABC$, point $D$ is chosen on side $AB$ and point $E$ is chosen on side $AC$ so that $DE$ is parallel to $BC$ and $\frac{BC}{DE} = \sqrt{2}$. The process is repeated two more times. Point $F$ is chosen on $CA$ and point $G$ on $BA$ so that $FG$ is parallel to $CA$ and $\frac{CA}{FG} = \sqrt{2}$. Point $H$ is chosen on side $CB$ and $I$ on $CA$ so that $HI$ is parallel to $BA$ and $\frac{BA}{HI} = \sqrt{2}$. The segments $FG$ and $HI$ intersect at $X$, $DE$ and $HI$ intersect at $Y$, and $DE$ and $FG$ intersect at $Z$.

SSCE HSC Mathematics Extension 2 Applications of mathematical induction: Use mathematical induction to pr

Use mathematical induction to prove that, for $n \geq 1$, $$x^{(3n)} - 1 = \left( x - 1 \right) \left( x^{2n} + x^{n+1} \right) \cdots \left( x^{(n-1)(2n+1)} + x^{(n-1)(3)} + 1 \right)$$ In $\triangle ABC$, point $D$ is chosen on side $AB$ and point $E$ is chosen on side $AC$ so that $DE$ is parallel to $BC$ and $\frac{BC}{DE} = \sqrt{2}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2018 - Paper 1

Question 16

$Use-mathematical-induction-to-prove-that,-for-$n-\geq-1$,--$$x^{(3n)}---1-=-\left(-x---1-\right)-\left(-x^{2n}-+-x^{n+1}-\right)-\cdots-\left(-x^{(n-1)(2n+1)}-+-x^{(n-1)(3)}-+-1-\right)$$--In-$\triangle-ABC$,-point-$D$-is-chosen-on-side-$AB$-and-point-$E$-is-chosen-on-side-$AC$-so-that-$DE$-is-parallel-to-$BC$-and-$\frac{BC}{DE}-=-\sqrt{2}$-HSC-SSCE Mathematics Extension 2-Question 16-2018-Paper 1.png$

Use mathematical induction to prove that, for $n \geq 1$, $$x^{(3n)} - 1 = \left( x - 1 \right) \left( x^{2n} + x^{n+1} \right) \cdots \left( x^{(n-1)(2n+1)} + x^{(... show full transcript

Worked Solution & Example Answer:Use mathematical induction to prove that, for $n \geq 1$, $$x^{(3n)} - 1 = \left( x - 1 \right) \left( x^{2n} + x^{n+1} \right) \cdots \left( x^{(n-1)(2n+1)} + x^{(n-1)(3)} + 1 \right)$$ In $\triangle ABC$, point $D$ is chosen on side $AB$ and point $E$ is chosen on side $AC$ so that $DE$ is parallel to $BC$ and $\frac{BC}{DE} = \sqrt{2}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2018 - Paper 1

Step 1

(i) Prove that $DY = ZE$.

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Answer

In triangles $BGC$ and $BAC$ , since $FG \parallel CA$ and $DE \parallel BC$ , we can apply the Basic Proportionality Theorem (Thales' theorem).

Since:

$\angle BGC = \angle BAC$ (corresponding angles)
$\angle GBC = \angle CBA$ (alternate angles)

Thus, triangles $BGC \sim BAC$ . This similarity gives us the following relationships:

$\frac{BC}{CA} = \frac{BG}{BA} = \frac{GC}{AC}$

Moreover, since $DE \parallel BC$ and using the same theorem, we get:

$\frac{DY}{ZE} = \frac{BC}{AC}$

Therefore, since $DE$ is a transversal cutting $BC$ and $AC$ , we can conclude that:

$DY = ZE$

Step 2

(ii) Find the exact value of the ratio $\frac{YZ}{BC}$.

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Answer

Considering triangles $ZDE$ and $CBF$ , we can observe:

$\frac{DE}{BC} = \frac{YZ}{BC}$

From part (i), we have established that $DE = DY + ZE$ . Given that $\frac{BC}{DE} = \sqrt{2}$ from the problem statement, we also know:

$YZ = DE - DY$ .

Substituting gives:

$YZ = \frac{BC}{\sqrt{2}} - DY$

From our previous conclusions, knowing that $DY = ZE$ implies that:

$YZ = 2ZE$ .

Finally, substituting back yields:

$\frac{YZ}{BC} = \frac{2ZE}{BC} = 2 \cdot \frac{ZE}{BC}$ .

(i) Prove that $DY = ZE$.

(ii) Find the exact value of the ratio $\frac{YZ}{BC}$.

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