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