December 09, 2019, 11:23:15 am

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#### Jefferson

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##### Binomial Theorem
« on: June 05, 2019, 07:11:44 pm »
0
2012 HSC Q11 (f) ii. (see attachment 1)

General Term = nCk (2x3) k (-x -1) n-k
x0 for constant term, i.e.
0 = 3k - (n-k)
0 = 4k - n
n = 4k         ∴ n must be a multiple of 4.

Approach 2 (Alternate way to represent general term, also on reference sheet):
General Term = nCk (2x3) n-k (-x-1) k
x0 for constant term, i.e
0 = 3(n-k) - k
0 = 3n - 3k - k
n = 4k/3

This also shows that 'n' must still be a multiple of 4. However, there is now a restriction on 'k' (i.e. it must be an integer multiple of 3 to make 'n' an integer") that did not exist in the first approach..
Why does this happen and what does it mean? How should we interpret the 'division by 3'?

This question could've easily been written back to front, i.e. (-1/x + 2x3) n), so sticking with the 1st approach is not an answer I'm looking for. What I'm after is how to interpret the 4k/3 rather than the solution to the question itself. Please clarify this for me.

Thank you!
« Last Edit: June 05, 2019, 07:21:21 pm by Jefferson »

#### RuiAce

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##### Re: Binomial Theorem
« Reply #1 on: June 05, 2019, 07:25:45 pm »
+3
2012 HSC Q11 (f) ii. (see attachment 1)

General Term = nCk (2x3) k (-x -1) n-k
x0 for constant term, i.e.
0 = 3k - (n-k)
0 = 4k - n
n = 4k         ∴ n must be a multiple of 4.

Approach 2 (Alternate way to represent general term, also on reference sheet):
General Term = nCk (2x3) n-k (-x-1) k
x0 for constant term, i.e
0 = 3(n-k) - k
0 = 3n - 3k - k
n = 4k/3

This also shows that 'n' must still be a multiple of 4. However, there is now a restriction on 'k' (i.e. it must be an integer multiple of 3 to make 'n' an integer").
Why does this happen? How should we interpret the 'division by 3'?

This question could've easily been written back to front, i.e. (-1/x + 2x3) n), so sticking with the 1st approach is not an answer I'm looking for. What I'm after is how to interpret the 4k/3 rather than the solution to the question itself. Please clarify this for me.

Thank you!
$k$ is the ‘index’ that gives you the constant term.

The idea is that the expression you wrote will represent the $k$-th term upon expanding that binomial term. But of course, we rig it so that the value of $k$ we require gives the constant term.

The answer you’re seeking is that you’ve proven something extra in the other representation. When you choose the general term to correspond to the alternate representation, the index $k$ that will give you the constant term must in fact be a multiple of 3.

You can try expanding things out. You should find that using the alternate representation, it will always be a “multiple of 3”-th term that gives you the constant term. (Note: Technically speaking, because $k$ is allowed to equal $0$, we have a “0-th” term, and then the 1st, 2nd, 3rd and so on-th term.)

#### Jefferson

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##### Re: Binomial Theorem
« Reply #2 on: June 05, 2019, 08:07:04 pm »
0
$k$ is the ‘index’ that gives you the constant term.

The idea is that the expression you wrote will represent the $k$-th term upon expanding that binomial term. But of course, we rig it so that the value of $k$ we require gives the constant term.

The answer you’re seeking is that you’ve proven something extra in the other representation. When you choose the general term to correspond to the alternate representation, the index $k$ that will give you the constant term must in fact be a multiple of 3.

You can try expanding things out. You should find that using the alternate representation, it will always be a “multiple of 3”-th term that gives you the constant term. (Note: Technically speaking, because $k$ is allowed to equal $0$, we have a “0-th” term, and then the 1st, 2nd, 3rd and so on-th term.)

Thanks for the clarification, RuiAce. That makes sense.
Is there a concrete reason as to why " Approach 1 " did not have this extra representation other than just the algebra at work? (since that seems a little arbitrary to me).

#### RuiAce

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##### Re: Binomial Theorem
« Reply #3 on: June 05, 2019, 08:58:03 pm »
+2
Thanks for the clarification, RuiAce. That makes sense.
Is there a concrete reason as to why " Approach 1 " did not have this extra representation other than just the algebra at work? (since that seems a little arbitrary to me).

The subtlety behind that is actually more or less in what happens when we (sort-of) "flip" numbers in a number pyramid. (Note of course that this is purely for investigation, and not really a part of the course.)

Constant terms in the expansion of $\left(2x^3-\frac1x\right)^n$ only appear when $n$ is a multiple of $4$ as already established, so I'll only consider multiples of 4 in this analogy. The idea is that:
- When $n=0$, we will have 1 term. It can just be indexed by 0.
- When $n=4$, we will have 5 terms. We can index them by 0, 1, 2, 3, 4.
- When $n=8$, we will have 9 terms. We can index them by 0, 1, 2, 3, 4, 5, 6, 7, 8
... and so on.

In each approach, a particular index $k$ will yield the constant term. This index will be bolded for each approach. But what happens is this:

Approach 1:
n=0  : 0
n=4  : 0 1 2 3 4
n=8  : 0 1 2 3 4 5 6 7 8
n=12: 0 1 2 3 4 5 6 7 8 9 10 11 12
n=16: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Approach 2:
n=0  : 0
n=4  : 0 1 2 3 4
n=8  : 0 1 2 3 4 5 6 7 8
n=12: 0 1 2 3 4 5 6 7 8 9 10 11 12
n=16: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

We see that in approach 1, as $n$ increases by 4, the correct index $k$ increases by 1. Whereas in approach 2, as $n$ increases by 4, the correct index $k$ increases by 3. This completes the illustration of how the two approaches differ. (Note also that approach 1 will, in theory, hit every single integer.)

To top off the why: The why relies on how when we go from approach 1 to approach 2. What we're actually doing when we use the other formula on the reference sheet is that we're saying to reverse all of the terms. So for example when $n=8$:
- The 0th term became the 8th term
- The 1st term became the 7th term
- The 2nd term became the 6th term - this is the culprit behind why 6 is bolded for n=8
- The 3rd term became the 5th term
...and so on forth. You can probably use some reasonably simple arithmetic progressions (common difference = 1) to investigate why the increments then change from 1 to 3.

#### Jefferson

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##### Re: Binomial Theorem
« Reply #4 on: June 10, 2019, 10:58:46 am »
+1