What's the meaning behind a coefficient of $x^r$ in case of $(1-x)^{-n}$? If we r not given any condition on x .

Question

Like suppose we have to find coefficient of $x^n$ in the expansion of $[(1+x)^{3n+1}]\cdot (1-x)^{-2}$, why we assume the $|x|$ less than $1$ and expand that to get the coeffient, why not we divide that $(1+x)^{3n+1}$ polynomial from $(1-x)^2$ to get the resulting polynomial which will have a coefficient of $x^n$ in it ? Will this not work?

Answer 1

Let's look at $$ \frac{(1+x)^{3n+1}}{(1-x)^{2}} $$ for a moment. This is a rational function. It is undefined at $x = 1$ (because division by zero is undefined).

The binomial theorem allows us to replace $(1+x)^{3n+1}$ with \begin{align*} (1+x)^{3n+1} &= \sum_{k=0}^{3n+1} \binom{3n+1}{k} x^k \\ &= 1 + (3n+1)x + \frac{3}{2}n(3n+1)x^2 + \cdots \end{align*} and (using Newton's form) also $(1-x)^{-2}$ with \begin{align*} (1-x)^{-2} &= \sum_{j=0}^{\infty} \binom{-2}{j} x^j \\ &= 1 + 2x + 3x^2 + \cdots \text{.} \end{align*}

The first series converges for all $x$ for any choice of $n$. In fact, the first series is just a polynomial of degree $3n+1$. The second series, expanded around $x = 0$, has radius of convergence $1$, since the asymptote at $x = 1$ is the closest singularity of the function (in the complex plane). Consequently, for that series to be valid, $|x| < 1$.

If we want the coefficient of $x^{50}$ in the product of these two series, we can truncate each series to degree $50$ (because higher degree terms will not reduce their degree when multiplied by terms in the other series) and perform the multiplication. The first few terms are $$ (1 + (3n+1)x + \frac{3}{2}n(3n+1)x^2 + \cdots) \cdot (1 + 2x + 3x^2 + \cdots) = 1 + (3n+3)x + \frac{9n^2 + 15n + 10}{2}x^2 + \cdots \text{.} $$ Continuing to degree $50$, we get a coefficient of $x^{50}$ that is a degree $50$ polynomial in $n$. Note that one series is only valid for $|x| < 1$, so this product is only valid for $|x| < 1$, so this coefficient is only valid for $|x| < 1$.

Answer 2

You actually don't have to assume $|x| < 1$ since you can treat the expressions as "formal" power series in $x$.

At any rate, by the (extended) binomial theorem, $$ (1+x)^{3n+1} = \sum_{k\geq 0}{\binom{3n+1}{k}}x^k \\ (1-x)^{-2} = \sum_{j\geq 0}{jx^j} $$ then the coefficient of $x^n$ is $$ [x^n] \left( (1-x)^{-2}(1+x)^{3n+1} \right) = \sum_{j+k = n} j\binom{3n+1}{k} \\ = \sum_{k} (n-k)\binom{3n+1}{k} $$ That sum can be broken into two terms, the first is just the constant $n$ times the sum of binomial coeffcients $\binom{3n+1}{k}$ and the second is a sum of $kbinom{3n+1}{k}$. Neither of these is trivial, as both are partial sums of binomial coefficients. I suspect, however, that a closed-form answer exists in terms of Bernoulli numbers.