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March 31, 2020, 12:28:44 pm

Author Topic: Need help with these induction questions  (Read 179 times)  Share 

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Benicillin

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Need help with these induction questions
« on: February 15, 2020, 12:14:19 am »
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Need help thanks.

fun_jirachi

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Re: Need help with these induction questions
« Reply #1 on: February 15, 2020, 02:16:03 pm »
+6
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! :)



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! :)

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Benicillin

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Re: Need help with these induction questions
« Reply #2 on: February 16, 2020, 08:28:26 pm »
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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.

fun_jirachi

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Re: Need help with these induction questions
« Reply #3 on: February 22, 2020, 10:39:18 pm »
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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


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


Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Advanced [87] | Maths Extension 1 [98] | Maths Extension 2 [97]
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning [740] | Decision Making [890] | Quantitative Reasoning [880] | Abstract Reasoning [800]