Deriving Maclaurin series

Consider the convergent geometric series \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\) where&nbsp;\(|r| &lt; 1\).
If we let&nbsp;\(a = 1\) and \(r = x\), then we have&nbsp;\(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) where&nbsp;\(|x| &lt; 1\).
We can represent the function \(\mathrm{f}(x) = \frac{1}{1-x}\) where&nbsp;\(|x| &lt; 1\) with the power series \[
\sum_{n=0}^{\infty} x^n,
\] where&nbsp;\(|x| &lt; 1\).

Consider the convergent geometric series \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\) where \(|r| < 1\).

If we let \(a = 1\) and \(r = x\), then we have \(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) where \(|x| < 1\).

We can represent the function \(\mathrm{f}(x) = \frac{1}{1-x}\) where \(|x| < 1\) with the power series 
\[
\sum_{n=0}^{\infty} x^n,
\] 
where \(|x| < 1\).

Binomial Expansion as a Maclaurin Series

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International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Motivating example

Motivating example