Answer
$ \dfrac{1}{(1-x)^3}=1+3x +6x^2+10x^3+15x^4...........$
Work Step by Step
The Taylor`s Series can be represented as: : $T(x)=\Sigma_{n = 0}^{\infty} \dfrac{f^n (c)}{n!}(x-c)^n$
If we consider $c=0$, then Taylor`s Series, $T(x)$ is also called as Maclaurin Series and the Maclaurin series for $f(x)$ is as given below:
$f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+..............$
From Table $2$, we write the Maclaurin series for $f(x)=\dfrac{1}{1-x}=\Sigma_{n = 1}^{n}x^n=1+x+x^2+x^3+.....$
Here, we will differentiate $f(x)=\dfrac{1}{1-x}$ with respect to $x$.
So, $\dfrac{d}{dx}(\dfrac{1}{1-x})=\dfrac{d}{dx}(1+x+x^2+x^3+....)$
or, $\dfrac{1}{(1-x)^2}=1+2x+3x^2+4x^3+5x^4+...... ~~~~~(1)$
Now, we will again differentiate the equation (1) with respect to $x$.
$\dfrac{d}{dx}(\dfrac{1}{(1-x)^2})=\dfrac{d}{dx}(1+2x+3x^2+4x^3+5x^4+...... )$
or, $(-1) \times \dfrac{-2}{(1-x)^3}=2+6x +12x^2+20x^3+.....$
Thus, we get: $ \dfrac{1}{(1-x)^3}=1+3x +6x^2+10x^3+15x^4...........$
