Power Series Expansion of Cube Root of 1 + x

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Theorem

Let $x \in \R$ such that $-1 < x \le 1$.

Then:

$\sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots$


Proof

\(\ds \sqrt [3] {1 + x}\) \(=\) \(\ds \paren {1 + x}^{\frac 1 3}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {\frac 1 3}^{\underline k} } {k!} x^k\) General Binomial Theorem
\(\ds \) \(=\) \(\ds 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\frac 1 3 - j} } {k!} x^k\) Definition of Falling Factorial and extracting $k = 0$
\(\ds \) \(=\) \(\ds 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {1 - 3 j} } {3^k k!} x^k\) simplifying
\(\ds \) \(=\) \(\ds \sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots\)

$\blacksquare$


Sources