binomial coefficient of x

Question

I have some questions that are bothering me for a long time and couldn't find an answer online, so I'll try to make it short.

In any binomial expansion, for instance of the polynomial $P(a, x, n) = (1 +ax)^n$:

1. Is the sum of $x$'s coefficients in even indexes equal to the sum of $x$'s coefficients in odd indexes?

1.a If the above is true, and the proof to it is:

Consider $T(x) = (1+x)^2$, set $x = -1$, then we have $T(-1) = 0$ and $x$'s coefficients in odd numbers become negative, we move them to the other side, which was before 0 and then even $x$ coefficients in odd powers are equal in sum to $x$'s coefficients in even powers shown exatcly here: (https://www.youtube.com/watch?v=DRpYdq4EmNM)

The problem I have with this proof is: what happens if you start with $T(x) = (1+x^2)^2$?

Another problem is: let's assume we have a polynomial with 49 odd indexed coefficients and 50 even indexed ones (as in $P(-1, 1, 98)$, for instance), and then lets just say I change $n$ to 99; now we have 50 terms in each sum, but the number of terms of even index doesn't change, while the number of odd indexed ones did change. Are the sums still equal?

2. About calculating the sums of even and odd powers.

The sum of $x$'s coefficients in total of $P=(1+ax^2)^n$ is $(1+1^2)^n$ (setting $x = 1, a = 1$).

According to 1. above, this equals to the sum of $x$ coefficients in odd powers and the sum of $x$ coefficients in even powers, and they both are equal, so according to this, to get either the even or odd powers sum, can I just divide the total sum by 2? One of the "official" ways to get the sum $S$ of $x$'s coefficients in the even powers is $S = (P(1)+P(-1))/2$. Now, I do understand this method but just to be clear: in $(P(1)+P(-1))$ we eliminate all coefficients of $x^k$ for $k$ odd and then we're left with only $x$'s coefficients in even indexes multiplied by two, so to get $S$ I have to divide by 2, is this correct?

Sorry it's not as short as needed but its much shorter than it was orignially; thanks in advance and sorry for bad english.

As it stands, this is unreadable. Please edit for clarity. I suggest asking one question at a time rather than running on like this. — lulu, Sep 08 '19 at 20:29
I expect this question will be closed soon because, in its current form, it is incoherent. Please try to edit it. What is the question? You didn't tell us what you are looking at. What is $X$? What are $X's$ coefficients? We can't read your mind...give us something to work with. — lulu, Sep 08 '19 at 20:34
Voting to close the question. If you can, please edit for clarity. — lulu, Sep 08 '19 at 20:49
i tried editing it to make it better, sorry, first time in the site — avi123123, Sep 08 '19 at 20:51
The edited question makes no sense. If you take $a=2$, $n=1$ then you have $2x+1$ so obviously the coefficients are not equal. If you have a better question, please ask it. — lulu, Sep 08 '19 at 22:46

Jose Ramirez · Accepted Answer · 2019-09-08T22:11:21.210

0

Well, I believe I 'understood' the question.

In general, for $P(a, x, n) = (1 + ax)^n$ it doesn't hold that the sum of the coefficients of the odd powers of $x$ equals the sum of the coefficients of the even powers of $x$.

In particular, setting $a = 1$ in $P$, expanding by the binomial theorem, and rearranging after setting $x = -1$ (since we would then have that $P(1, -1, n) = (1 - 1)^n = 0$), we get the main result shown in the post's video, i.e., ${n \choose 1} + {n \choose 3} + \cdots = {n \choose 0} + {n \choose 2} + \cdots$

Note that the value of $n$ is irrelevant for the proof you mention to work, like, if you change $n$, both sums of odd and even coefficients will change, even though the number of terms might stay the same for one sum or the other.

Now, by starting with the polynomial $(1 + x^2)^n$ you can't get the same result as the one in the video.

(P(1)+P(-1))/2. now i do understand this method but just to be clear this method eliminates all ODD powers coefficients and then you have only X's coeffieints in even indexes times two so you divide by 2, is this correct?

For polynomials in one variable, yeah, this is correct.

edited Sep 08 '19 at 22:11

answered Sep 08 '19 at 22:06

Jose Ramirez

343

first off thanks, and let me just see if i got you right "it doesn't hold that the sum of the coefficients of the odd powers of x equals the sum of the coefficients of the even powers of x." my english is bad, does that mean that X's coefficients in even powers does or does not equal to X coefficients in odd powers? or is it true only when you set 1 and then -1 (what happens if the result with -1 isnt 0?) – avi123123 Sep 09 '19 at 04:25
well, the sums of the coefficients are not always equal; set $a = 2, n = 3$ in $(1 + ax)^n$; by expanding, we get $(1 + 2x)^3 = 1 + 6x + 12x^2 + 8x^3$; the sum of the odd indexed coefficients is 14, and of the evens, 13. Is it only true in the case I described above? Good question, but the thing is that by finding a case where it doesn't hold you can safely say that the sums aren't always equal. The fact that there are cases where the polynomial I described above can be identical to 0 is what makes the proof of the video work; if it isn't, then you'd be trying to prove something else entirely. – Jose Ramirez Sep 09 '19 at 10:26
i see! maybe the pattern is that x=-1 will reset the equatio, anyway, let me ask you this, when i can prove that x's evens sum equals to x's odd sum, is it currect to assume that i can get the odd sum or the even sum by getting the overall sum (which is setting X=1) divided by 2? – avi123123 Sep 09 '19 at 11:12
If both are the same, yep, you can – Jose Ramirez Sep 09 '19 at 12:09
alright i think i got it, thanks just one more last thing, why i think it wont work on anything that isnt 1,-1 "combo" if you can just approve that lets say what resets the equation is -2 for instance (2-x)^n, when trying to get the same proof equation, it will be entirely different values between the sum of even powers and odd powers, right? there wont be any relation between the sums? – avi123123 Sep 09 '19 at 13:14
Well, in that particular case you describe, no; you would basically get the same sums of the video multiplied by $2^n$; in that case, you cancel out that power and you get the sums back. Every set of values you plug in the polynomial you start with will create a relation in the end; the question in that case is: is this the relation I want to prove? – Jose Ramirez Sep 09 '19 at 14:32
i see, and those relations would be equal? in (x+a)^n as long as x=-a and i set out the sums they will be equal? (x-2)^n, the odd and even indexes are not equal unless you set x=-2 but in cases in which x=-1 will RESET the equation the odd and even coefficients sums will always be equal for example (2x^2 +x -1)^3, how can you explain that? – avi123123 Sep 09 '19 at 15:23
yes, for $x = -a$ both sides of whatever you get would be equal, since for that value of $x$ they add up to 0. In your second case it's the same idea, i.e., for $x=1/2$ or $x=-1$ you would have a bunch of terms that add up to zero, rearranging them by index would give you a relation as the one you seek. – Jose Ramirez Sep 09 '19 at 15:41
yes what i meant was when x=-1 the sum of even and odd places coeffieints will ALWAYS be equal even if you dont put -1 while when its other than that, eg -2 it wont be equal unless you set x=-2, this confused me, untill i realized that in the first case setting x=-1 is very close to not setting any x or just setting x=1 anyway, thank you very much for your patience, as i am new i dont know how to give credit or mark answers, i think i upvoted your answer, if theres anything more i can do please tell – avi123123 Sep 09 '19 at 15:55
you can't say always, because it will depend on the polynomial you start with; for instance, you get what you say with $x=1/2$ as well with $(2x^2+x-1)^3$. $(1+ x)^n$ is just by coincidence a proper choice to prove the equality. – Jose Ramirez Sep 09 '19 at 16:16
and to mark as accepted, just click on the check mark at the left of the answer :) – Jose Ramirez Sep 09 '19 at 16:25

binomial coefficient of x

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