The \(n\)th derivative of a function \(f\) at \(x=0\) is given by \(f^{(n)}(0)= (-1)^n \dfrac{n+1}{(n+2)2^n} \) for all \(n \geq 0\). Which of the following is the Maclaurin series for \(f\) ?

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The terms of the Maclaurin polynomial is given by:

$$ f^{(n)}(0) \cdot \frac{x^n}{n!} $$ $$ (-1)^n \dfrac{n+1}{(n+2)2^n} \cdot \frac{x^n}{n!} $$

The first three terms:

$$ (-1)^0 \dfrac{0+1}{(0+2)2^0} \frac{x^0}{0!} + (-1)^1 \dfrac{1+1}{(1+2)2^1} \frac{x^1}{1!} + (-1)^2 \dfrac{2+1}{(2+2)2^2} \frac{x^2}{2!} $$ $$ = \frac{1}{2}-\frac{1}{3}x + \frac{3}{32}x^2 $$