Finding coefficients of a polynomial expansion

Question

I am trying to find the coefficient of $x^{n}$ where $n \in[0, \infty)$ in the expansion

$(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 ... +x^{\infty}$)$^{t}$, where $t$ is an integer.

I haven't been able to distinguish a pattern between the coefficients and the value of $t$, but the sum $\sum_{i = 0}^{\infty} x^i$ is the Taylor series expansion of the geometric sum formula. I also tried some things with the binomial theorem, but was not able to make too much progress.

Answer 1

Assuming $|x| < 1$ use GP sum and the expression reduces to,

$$(1-x)^{-t}$$

Coefficient of $x^n$ in $(1-x)^{-t}$ equals $\binom{n+t-1}{t-1} = \binom{n+t-1}{n}$.

Check out the binomial expansion of $(1+x)^m$ when $m$ is negative.

Answer 2

$\left(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 ... +x^{\infty}\right)^t=\dfrac{1}{(1-x)^t}=(1-x)^{-t}$

The binomial expansion gives $$(1-x)^{-t}=\sum _{n=0}^{\infty } (-1)^n x^n \binom{-t}{n}$$ For $t=10$ for instance we have $$\left(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 ... +x^{\infty}\right)^{10}=1 + 10 x + 55 x^2 + 220 x^3 + 715 x^4 + 2002 x^5 + 5005 x^6 + 11440 x^7 + 24310 x^8 + 48620 x^9 + 92378 x^{10}+\ldots$$