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.