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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: Benicillin on February 15, 2020, 12:14:19 am

Title: Need help with these induction questions
Post by: Benicillin on February 15, 2020, 12:14:19 am: Need help thanks.
Title: Re: Need help with these induction questions
Post by: fun_jirachi on February 15, 2020, 02:16:03 pm: Hey there!

I'll demonstrate the first one, which requires the most effort to do, but you should really have a second attempt at the other two - please show us what you've done so we can help you improve and learn! :)

$\text{From part (a), we know that } \sin(A+B) + \sin(A-B) = 2\sin A\cos B \\ \text{Now, for } n=1 \\ \text{LHS } = \sin x, \ \text{RHS } = \frac{\sin\left(\frac{(1)x}{2}\right)\times\sin\left(\frac{((1)+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \\ \text{ie. LHS = RHS} \\ \text{Assume true for } n = k, \ k \in \mathbb{Z} ^+ \\ \text{ie. } \sin x + \sin 2x + \sin 3x + ... + \sin kx = \frac{\sin\left(\frac{kx}{2}\right)\times\sin\left(\frac{(k+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \\ \text{Now, proving true for } n = k+1, \\ \text{ie. } \sin x + \sin 2x + \sin 3x + ... + \sin kx + \sin (k+1)x = \frac{\sin\left(\frac{(k+1)x}{2}\right)\times\sin\left(\frac{(k+2)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \\ \begin{align*} \text{LHS } &= \frac{\sin\left(\frac{kx}{2}\right)\times\sin\left(\frac{(k+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} + \sin (k+1)x \ \ \text{(from assumption)} \\&= \frac{\sin\left(\frac{kx}{2}\right)\times\sin\left(\frac{(k+1)x}{2}\right) + \sin (k+1)x\times\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \\&= \frac{\sin\left(\frac{kx}{2}\right)\times\sin\left(\frac{(k+1)x}{2}\right) + 2\sin\left(\frac{(k+1)x}{2}\right)\cos\left(\frac{(k+1)x}{2}\right)\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \ \text{(using the sine double angle)} \\&= \frac{\sin\left(\frac{kx}{2}\right)\times\sin\left(\frac{(k+1)x}{2}\right) + \sin\left(\frac{(k+1)x}{2}\right)\left(\sin\left(\frac{(k+2)x}{2}\right) + \sin\left(\frac{-kx}{2}\right)\right)}{\sin\left(\frac{x}{2}\right)} \ \text{(Using the result from part (a))} \\&= \frac{\sin\left(\frac{(k+1)x}{2}\right)\times\sin\left(\frac{(k+2)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \\&= \text{RHS} \end{align*}$

A few things to note:
- Firstly, look at what you're working towards -> assume that in your working, that what you've got is correct and equivalent to the result (it's a matter of how you get there), and look ahead for tips on how to do that, especially through results that were given to you previously
- It's also handy to look out for other ways to integrate previous results (like how we used the sine double angle to get a cosine in the working to use part (a))

Hope this helps! :)
Title: Re: Need help with these induction questions
Post by: Benicillin on February 16, 2020, 08:28:26 pm: Thanks for that. I have been trying to work on question 6 (x^n - y^n / y - n) but the long line of xs and ys keep catching me off guard. Also with the one about being divisible by 6, Im a bit stuck because after factoring the equation of n = k+ 1, 6 was not a common factor.
Title: Re: Need help with these induction questions
Post by: fun_jirachi on February 22, 2020, 10:39:18 pm: Quote from: Benicillin on February 16, 2020, 08:28:26 pm
Thanks for that. I have been trying to work on question 6 (x^n - y^n / y - n) but the long line of xs and ys keep catching me off guard. Also with the one about being divisible by 6, Im a bit stuck because after factoring the equation of n = k+ 1, 6 was not a common factor.

Sorry for the late reply!

With the divisibility question, I'm going to put the answer up, but I'm going to give a few hints for Q6.

Skipping the first step, since I'm sure you've done that
$\text{Assume that } k^3-k=6a \ \ (a,k \in \mathbb{Z}^+) \\ \text{Now, prove that } (k+1)^3-(k+1)=6b \ \ (b \in \mathbb{Z}^+) \\ \begin{align*} \text{LHS } &= (k+1)^3-(k+1) \\&= k^3+3k^2+2k \\&= k^3-k+3k^2+3k \\&= 6a+3k(k+1) \ \ \text{(from assumption)} \end{align*}$

I think it's safe to assume you got up to here, then realised hang on, that's a 3, not a 6. But consider k(k+1) for a second. In any two consecutive integers (we gave the condition that k was an integer!) one must be even, the other must be odd. That means we can actually express two consecutive integers as 2c(2c+1) for some integer c, or if k was odd, (2c+1)(2c+2) for some integer c. Essentially, what this all means is that k(k+1) must be even, and thus 3k(k+1) is an integer divisible by 6. Does this make sense? I think it's relatively straightforward from here since we now have some integer multiplied by 6 :)

For Q6, here's the hint (which basically gives away the answer, so look at your own discretion):
Spoiler
$\frac{x^{n+1}-y^{n+1}}{x-y} = x^n+\frac{y(x^{n}-y^{n})}{x-y}$