After working out binomial expansion of $\frac{1-x}{10}$ to power of $-3$ for my homework, I have to determine the value of $\frac{1}{0.999}$ to power of $3$ correct to $14$ decimal places. We haven't done it in lessons yet and the textbook doesn't make sense.
2026-03-31 23:25:27.1774999527
Binomial Series Expansion
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\begin{align} \pars{1 - {x \over 10}}^{-3} & = \sum_{n = 0}^{\infty}{-3 \choose n}\pars{-\,{x \over 10}}^{n} = \sum_{n = 0}^{\infty}\bracks{{n + 2 \choose n}\pars{-1}^{n}} \bracks{{\pars{-1}^{n} \over 10^{n}}x^{n}} \\[5mm] & = \sum_{n = 0}^{\infty} {\pars{n + 2}\pars{n + 1} \over 2 \times 10^{n}}\,x^{n} \end{align}
\begin{align} \pars{1 \over 0.999}^{3} & = 0.999^{\,-3} = \pars{1 - 0.001}^{-3} = \pars{1 - {0.01 \over 10}}^{-3} = \sum_{n = 0}^{\infty} {\pars{n + 2}\pars{n + 1} \over 2 \times 10^{n}}\,\pars{0.01}^{n} \\[5mm] & = \sum_{n = 0}^{\infty} {\pars{n + 2}\pars{n + 1} \over 2 \times 1000^{n}} \end{align}
$$ \sum_{n = 0}^{\color{#f00}{N}} {\pars{n + 2}\pars{n + 1} \over 2 \times 1000^{n}}\qquad \implies \qquad \begin{array}{cl}\hline \ds{\color{#f00}{N}} & \ds{\color{#f00}{\phantom{123456}value}} \\ \ds{0} & \ds{1} \\ \ds{1} & \ds{1.003} \\ \ds{2} & \ds{1.0030059999999998} \\ \ds{3} & \ds{1.0030060099999998} \\ \ds{4} & \ds{1.0030060100149998} \\ \ds{\large 5} & \ds{\color{#f00}{1.00300601001502}09} \\ \ds{6} & \ds{\color{#f00}{1.00300601001502}09} \\ \hline \end{array} $$