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Geometric type sums

Partial geometric sum:

for $|x|<1.$

Using the result for geometric sums similar results can be derived. Since

\displaystyle(1-x)^{-1}=1+x+x^{2}+\ldots=\sum_{i=0}^{\infty}x^{i}

then by differentiating, term by term, each side with respect to ${x}$ gives

	$\displaystyle(1-x)^{-2}$	$\displaystyle=$	$\displaystyle 1+2x+3x^{2}+\ldots$
		$\displaystyle=$	$\displaystyle\sum_{i=1}^{\infty}ix^{i-1}$

	$\displaystyle 2(1-x)^{-3}$	$\displaystyle=$	$\displaystyle 2+6x+12x^{2}+\ldots$
		$\displaystyle=$	$\displaystyle\sum_{i=2}^{\infty}i(i-1)x^{i-2}$