Not long ago, I derived the formula for Chebyshev polynomials
$$T_{n}\left( x\right)= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( x^2-1\right)^{k}$$ How to extract the coefficients of this polynomial of degree $n$ ?
I tried using Newton's binomial but got a double sum $$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( \sum_{m=0}^{k} {k \choose m} x^{2m}\left( -1\right)^{k-m} \right) \\ \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\sum_{m=0}^{k}\left( -1\right)^{k-m} {n \choose 2k} \cdot {k \choose m} x^{n+2m-2k}\\ $$
Now how to continue counting this sum ? What would it look like to change the order of summation and would it do anything ?
What else did I try ?
Well, I worked out the sum $$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( x^2-1\right)^{k} $$
for $n=8$ and I hypothesized that $$T_{n}\left( x\right)= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \sum_{m=k}^{\lfloor \frac{n}{2} \rfloor}\left( -1\right)^{k}{n \choose 2m} \cdot {m \choose k} x^{n-2k} $$ However, it would be useful to demonstrate the correctness of this hypothesis and count the sum of $$\sum_{m=k}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2m} \cdot {m \choose k}$$ Here I would like to point out that Wolfram Alpha counts this sum incorrectly $$\sum_{m=k}^{\lfloor \frac{n}{2}\rfloor}{{n \choose 2m} \cdot {m \choose k}} = \frac{n}{2n-2k} \cdot 2^{n-2k} \cdot {n - k \choose k}$$
And it would even be a nice result but first of all it is not quite correct ( Have you noticed why ?) and secondly it comes from a hypothesis I made after dissecting the formula for $n=8$.
It seems to me that this hypothesis of mine would be enough to prove by induction after n but how would it look?
How to calculate the coefficients of the Chebyshev polynomial ? Are you sure they will be the sum of products of Bernoulli numbers and Newton symbol since, after working out the formula, I came out that it would be $$T_{n}\left( x\right)= \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \sum_{m=k}^{\lfloor\frac{n}{2}\rfloor}\left( -1\right)^{k} {n \choose 2m} \cdot {m \choose k} x^{n-2k}$$
Please help.