Author Topic: VCE Specialist 3/4 Question Thread! (Read 2826744 times) Tweet Share

lzxnl · « **Reply #9210 on:** February 26, 2018, 10:58:51 pm »

Quote from: RuiAce on February 26, 2018, 04:19:48 pm

It's actually the standard method in HSC MX1. I still don't understand why they don't teach it in the VCE tbh.

Ironically I learnt that method in a HSC textbook about 11 years ago HAHAHAHAHAHAH
That book also taught using the tangent half-angle substitution to do this, although that seems overkill.
I don't get it either. It's the same logic as polar form of a complex number.

@noregret, there are two ways.
1. Formula bash. Arithmetic series with n = n, l = n, d = 1, giving you
$ S = \frac{n}{2}(n+1) $

2. A little bit of cunning (also known as deriving the formula.

S =
1 + 2 + 3 + 4... =
n + (n-1) + (n-2) + (n-3)...

There's n lots of terms. If we sum the above two vertically, can you see we get n lots of n+1?
$ \therefore 2S = n(n+1)\\ S = \frac{n}{2}(n+1) $

RuiAce · « **Reply #9211 on:** February 27, 2018, 12:31:36 am »

Quote from: lzxnl on February 26, 2018, 10:58:51 pm

Ironically I learnt that method in a HSC textbook about 11 years ago HAHAHAHAHAHAH

Nice to know that HSC math is brilliant

I think the main reason why they teach the tangent half-angle sub is because of 4U integrals of the form $ \int \frac{dx}{a + b\sin x + c \cos x} $, because without going into crafty integration techniques it's a nice bash. Of course, for definite integrals one would probably prefer complex analysis once they're at the level, but it's probably the better option in high school (especially since most integrals then are indefinite)

Although, I interpreted his sum as $1\times 2 + 3\times 4 + 5\times 6 + \dots$ - fairly sure that ain't an AP?

noregret · « **Reply #9212 on:** February 27, 2018, 03:19:03 pm »

Thanks for replyin my questions to both of you, my question is AP as RuiAce stated it.

JamesMaths · « **Reply #9213 on:** February 28, 2018, 01:47:06 pm »

Quote from: noregret on February 26, 2018, 08:53:19 pm

Hello, how do you solve

Find the sum to n terms (1 times 2)+(3 times 4)+(5 times 6)+... (factorise answer)?

Just wonder whether the questions is to calculate the
SUM of (1 times 2) + (2 times 3) + (3 times 4) + ... + (n times n + 1)
which can be shown by the induction method as
n(n + 1)(n + 2) / 3.

if it is (1 times 2) + (3 times 4) + (5 times 6) + ...
then it needs more efforts to link to the above result to solve the question.

noregret · « **Reply #9214 on:** February 28, 2018, 06:01:32 pm »

Quote from: JamesMaths on February 28, 2018, 01:47:06 pm

Just wonder whether the questions is to calculate the
SUM of (1 times 2) + (2 times 3) + (3 times 4) + ... + (n times n + 1)
which can be shown by the induction method as
n(n + 1)(n + 2) / 3. Thanks for your response.

if it is (1 times 2) + (3 times 4) + (5 times 6) + ...
then it needs more efforts to link to the above result to solve the question.

lzxnl · « **Reply #9215 on:** February 28, 2018, 06:10:34 pm »

Quote from: RuiAce on February 27, 2018, 12:31:36 am

Nice to know that HSC math is brilliant
I think the main reason why they teach the tangent half-angle sub is because of 4U integrals of the form $ \int \frac{dx}{a + b\sin x + c \cos x} $, because without going into crafty integration techniques it's a nice bash. Of course, for definite integrals one would probably prefer complex analysis once they're at the level, but it's probably the better option in high school (especially since most integrals then are indefinite)

Although, I interpreted his sum as $1\times 2 + 3\times 4 + 5\times 6 + \dots$ - fairly sure that ain't an AP?

LOL. I must have been on some crack when attempting to answer his question. That's awkward.

And you're talking about a very specific class of definite integral. What if I wanted to integrate from 0 to pi/3? Good luck using complex analysis for that.

Here's my favourite technique for doing sums like these.
$ S(n) = \sum_{k=1}^n k(k+1)\\ S(1) = 2\\ S(2) = 2 + 6 = 8\\ S(3) = 2 + 6 + 12 = 20\\ S(4) = 2 + 6 + 12 + 20 = 40\\ S(5) = 2 + 6 + 12 + 20 + 30 = 70 $
Here's a little known fact.
$ \sum_{k=1}^n p(k) $
is a mth order polynomial in n if p is a (m-1)th order polynomial. Then, here we know that p(k) = k(k+1), a quadratic, so we know the sum MUST be a cubic. Cubics are defined by four points, so all we need now is a cubic that satisfies S(1) = 2, S(2) = 8, S(3) = 20, S(4) = 40. You can show that the cubic given by a different poster is the only cubic that passes through these four points.

Another cool way:
$ S(n) = \sum_{k=1}^n k(k+1)\\ (k+1)^3 - k^3 = 3k(k+1) + 1\\ \therefore \sum_{k=1}^n (k+1)^3 - k^3 = 3\sum_{k=1}^n k(k+1) + \sum_{k=1}^n 1 = 3S(n) + n\\ (n+1)^3 - 1 = 3S(n) + n\\ S(n) = \frac{(n+1)^3 - (n+1)}{3} = \frac{(n+1)((n+1)^2 - 1)}{3} = \frac{n(n+1)(n+2)}{3} $

The first method always works, the second method is a cool trick that happens to work here. Note that I can evaluate the sum of the cubes because it telescopes; most of the terms in summing (k+1)^3 and k^3 will cancel, leaving only the two terms I've written.

noregret · « **Reply #9216 on:** February 28, 2018, 06:35:52 pm »

Quote from: lzxnl on February 28, 2018, 06:10:34 pm

LOL. I must have been on some crack when attempting to answer his question. That's awkward.

And you're talking about a very specific class of definite integral. What if I wanted to integrate from 0 to pi/3? Good luck using complex analysis for that.
Thanks for reply, it helped.
Here's my favourite technique for doing sums like these.
$ S(n) = \sum_{k=1}^n k(k+1)\\ S(1) = 2\\ S(2) = 2 + 6 = 8\\ S(3) = 2 + 6 + 12 = 20\\ S(4) = 2 + 6 + 12 + 20 = 40\\ S(5) = 2 + 6 + 12 + 20 + 30 = 70 $
Here's a little known fact.
$ \sum_{k=1}^n p(k) $
is a mth order polynomial in n if p is a (m-1)th order polynomial. Then, here we know that p(k) = k(k+1), a quadratic, so we know the sum MUST be a cubic. Cubics are defined by four points, so all we need now is a cubic that satisfies S(1) = 2, S(2) = 8, S(3) = 20, S(4) = 40. You can show that the cubic given by a different poster is the only cubic that passes through these four points.

Another cool way:
$ S(n) = \sum_{k=1}^n k(k+1)\\ (k+1)^3 - k^3 = 3k(k+1) + 1\\ \therefore \sum_{k=1}^n (k+1)^3 - k^3 = 3\sum_{k=1}^n k(k+1) + \sum_{k=1}^n 1 = 3S(n) + n\\ (n+1)^3 - 1 = 3S(n) + n\\ S(n) = \frac{(n+1)^3 - (n+1)}{3} = \frac{(n+1)((n+1)^2 - 1)}{3} = \frac{n(n+1)(n+2)}{3} $

The first method always works, the second method is a cool trick that happens to work here. Note that I can evaluate the sum of the cubes because it telescopes; most of the terms in summing (k+1)^3 and k^3 will cancel, leaving only the two terms I've written.

Mattjbr2 · « **Reply #9217 on:** February 28, 2018, 07:35:21 pm »

Can someone please tell me exactly what's wrong with method 1? I know that CAS shows that the range is [5,9], but still...

lzxnl · « **Reply #9218 on:** February 28, 2018, 07:37:21 pm »

Quote from: Mattjbr2 on February 28, 2018, 07:35:21 pm

Can someone please tell me exactly what's wrong with method 1? I know that CAS shows that the range is [5,9], but still...

The minimum value of f(x) + g(x) isn't necessary the sum of the minimum values. It's because they could attain their minimum values at different x values.

Mattjbr2 · « **Reply #9219 on:** February 28, 2018, 07:51:05 pm »

Quote from: lzxnl on February 28, 2018, 07:37:21 pm

The minimum value of f(x) + g(x) isn't necessary the sum of the minimum values. It's because they could attain their minimum values at different x values.

You're a legend.

noregret · « **Reply #9220 on:** February 28, 2018, 07:53:29 pm »

Hello, how you solve Sum (1^2+2^3)+(3^2+4^3)=(5^2+6^3)+....+ ((2n-1))^2+ (2n)^3)?

lzxnl · « **Reply #9221 on:** March 01, 2018, 12:42:06 pm »

Quote from: noregret on February 28, 2018, 07:53:29 pm

Hello, how you solve Sum (1^2+2^3)+(3^2+4^3)=(5^2+6^3)+....+ ((2n-1))^2+ (2n)^3)?

Well...you do the sums separately. I don't know where you'd get these questions, but anyway. We wish to first calculate the sum of the first n cubes and squares.
$ S(n) = \sum_{k=1}^n k^2\\ 3k^2 = (k+1)^3 - k^3 - 3k - 1\\ 3\sum_{k=1}^n k^2 = 3S(n) = \sum_{k=1}^n \left[(k+1)^3 - k^3 - 3k - 1\right]\\ = \sum_{k=1}^n \left[(k+1)^3 - k^3\right] - 3\sum_{k=1}^n k - n\\ = (n+1)^3 - 1 - \frac{3}{2}n(n+1) - n\\ = (n+1)\left((n+1)^2 - 1 - \frac{3}{2}n\right)\\ = (n+1)\left(n^2 + \frac{1}{2}n\right)\\ S(n) = \frac{n(n+1)(2n+1)}{6} $
where the sum involving the cubic terms telescopes (lots of terms cancel; check it yourself)
$ C(n) = \sum_{k=1}^n k^3\\ 4k^3 = (k+1)^4 - k^4 - 6k^2 - 4k^2 - 1\\ 4C(n) = \sum_{k=1}^n \left[(k+1)^4 - k^4 - 6k^2 - 4k - 1\right]\\ = \sum_{k=1}^n \left[(k+1)^4 - k^4\right] - 6\sum_{k=1}^n k^2 - 4\sum_{k=1}^n k - n\\ = (n+1)^4 - 1 - 6S(n) - 2n(n+1) - n\\ = (n+1)^4 - (n+1) - n(n+1)(2n+1) - 2n(n+1)\\ = (n+1)\left((n+1)^3 - 1 - n(2n+1) - 2n\right)\\ = (n+1)\left(n^3 + 3n^2 + 3n - 2n^2 - 3n\right)\\ = n^2(n+1)^2\\ C(n) = \frac{n^2(n+1)^2}{4} $
Your question can now be rephrased as
$ x = \sum_{k=1}^n \left[(2k-1)^2 + (2k)^3\right] $
The first term is the sum of all squares to 2n, minus the sum of the even squares.
$ \therefore x = S(2n) - \sum_{k=1}^n (2k)^2 + \sum_{k=1}^n 8k^3\\ = \frac{n(2n+1)(4n+1)}{3} - 4S(n) + 8C(n)\\ = \frac{n(2n+1)(4n+1)}{3} - \frac{2n(n+1)(2n+1)}{3} + 2n^2(n+1)^2\\ = \frac{n(2n+1)}{3}\left(4n+1 - 2(n+1)\right) + 2n^2(n+1)^2\\ = \frac{n(2n+1)(2n-1)}{3}+ 2n^2(n+1)^2\\ = \frac{n}{3}\left(4n^2 - 1 + 6n(n+1)^2\right)\\ = \frac{n}{3}\left(4n^2 - 1 + 6n^3 + 12n^2 + 6n\right)\\ = \frac{n}{3}\left(6n^3 + 16n^2 + 6n - 1\right) $

noregret · « **Reply #9222 on:** March 01, 2018, 05:17:22 pm »

Quote from: lzxnl on March 01, 2018, 12:42:06 pm

Well...you do the sums separately. I don't know where you'd get these questions, but anyway. We wish to first calculate the sum of the first n cubes and squares.
$ S(n) = \sum_{k=1}^n k^2\\ 3k^2 = (k+1)^3 - k^3 - 3k - 1\\ 3\sum_{k=1}^n k^2 = 3S(n) = \sum_{k=1}^n \left[(k+1)^3 - k^3 - 3k - 1\right]\\ = \sum_{k=1}^n \left[(k+1)^3 - k^3\right] - 3\sum_{k=1}^n k - n\\ = (n+1)^3 - 1 - \frac{3}{2}n(n+1) - n\\ = (n+1)\left((n+1)^2 - 1 - \frac{3}{2}n\right)\\ = (n+1)\left(n^2 + \frac{1}{2}n\right)\\ S(n) = \frac{n(n+1)(2n+1)}{6} $
where the sum involving the cubic terms telescopes (lots of terms cancel; check it yourself)
$ C(n) = \sum_{k=1}^n k^3\\ 4k^3 = (k+1)^4 - k^4 - 6k^2 - 4k^2 - 1\\ 4C(n) = \sum_{k=1}^n \left[(k+1)^4 - k^4 - 6k^2 - 4k - 1\right]\\ = \sum_{k=1}^n \left[(k+1)^4 - k^4\right] - 6\sum_{k=1}^n k^2 - 4\sum_{k=1}^n k - n\\ = (n+1)^4 - 1 - 6S(n) - 2n(n+1) - n\\ = (n+1)^4 - (n+1) - n(n+1)(2n+1) - 2n(n+1)\\ = (n+1)\left((n+1)^3 - 1 - n(2n+1) - 2n\right)\\ = (n+1)\left(n^3 + 3n^2 + 3n - 2n^2 - 3n\right)\\ = n^2(n+1)^2\\ C(n) = \frac{n^2(n+1)^2}{4} $
Your question can now be rephrased as
$ x = \sum_{k=1}^n \left[(2k-1)^2 + (2k)^3\right] $
The first term is the sum of all squares to 2n, minus the sum of the even squares.
$ \therefore x = S(2n) - \sum_{k=1}^n (2k)^2 + \sum_{k=1}^n 8k^3\\ = \frac{n(2n+1)(4n+1)}{3} - 4S(n) + 8C(n)\\ = \frac{n(2n+1)(4n+1)}{3} - \frac{2n(n+1)(2n+1)}{3} + 2n^2(n+1)^2\\ = \frac{n(2n+1)}{3}\left(4n+1 - 2(n+1)\right) + 2n^2(n+1)^2\\ = \frac{n(2n+1)(2n-1)}{3}+ 2n^2(n+1)^2\\ = \frac{n}{3}\left(4n^2 - 1 + 6n(n+1)^2\right)\\ = \frac{n}{3}\left(4n^2 - 1 + 6n^3 + 12n^2 + 6n\right)\\ = \frac{n}{3}\left(6n^3 + 16n^2 + 6n - 1\right) $

Thanks for your help, my questions come from books written by Steve Watson as I live in Tasmania we use these books as our school books.

lzxnl · « **Reply #9223 on:** March 02, 2018, 04:20:55 pm »

Quote from: noregret on March 01, 2018, 05:17:22 pm

Thanks for your help, my questions come from books written by Steve Watson as I live in Tasmania we use these books as our school books.

What topic is this from? Can I have a look at the other exercises in that section?

noregret · « **Reply #9224 on:** March 02, 2018, 08:44:11 pm »

So Steve Watson wrote Calculus, Sequences and Series, Complex Numbers, Matrices and Linear Transformations. My questions comes from Sequences and Series.

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2826744 times) Tweet Share

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