The coefficient of x^n, where n is any positi | vedclass

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Question

(A) Let $S = 1 + 2x + 3x^2 + \ldots \infty$. Multiplying by $x$, we get $xS = x + 2x^2 + 3x^3 + \ldots \infty$. Subtracting the two equations: $S(1-x)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $S = 1 + 2x + 3x^2 + \ldots \infty$. Multiplying by $x$, we get $xS = x + 2x^2 + 3x^3 + \ldots \infty$. Subtracting the two equations: $S(1-x) = 1 + x + x^2 + \ldots = \frac{1}{1-x}$. Thus, $S = \frac{1}{(1-x)^2} = (1-x)^{-2}$. The given expression is $(S)^{1/2} = ((1-x)^{-2})^{1/2} = (1-x)^{-1}$. Expanding $(1-x)^{-1}$ using the binomial theorem for negative indices: $(1-x)^{-1} = 1 + x + x^2 + \ldots + x^n + \ldots$. The coefficient of $x^n$ is $1$.

The coefficient of $x^n$, where $n$ is any positive integer, in the expansion of $(1+2x+3x^2+\ldots)^{-1/2}$ is

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