Power Series Expansion for Reciprocal of Cube of 1 + x
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Theorem
Let $x \in \R$ such that $-1 < x < 1$.
Then:
| \(\ds \dfrac 1 {\paren {1 + x}^3}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - 3 x + 6 x^2 - 10 x^3 + 15 x^4 - \cdots\) |
Proof 1
| \(\ds \frac 1 {\paren {1 + x}^2}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k\) | Power Series Expansion of $\dfrac 1 {\paren {1 + x}^2}$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac \d {\d x} \frac 1 {\paren {1 + x}^2}\) | \(=\) | \(\ds \frac \d {\d x} \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} x^k\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -\frac 2 {\paren {1 + x}^3}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 1} k x^{k - 1}\) | differentiating with respect to $x$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 2 {\paren {1 + x}^3}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \paren {k + 1} k x^{k - 1}\) | taking one of the $-1$s out | ||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^\infty \paren {-1}^{k - 1} \paren {k + 1} k x^{k - 1}\) | the term in $k = 0$ vanishes | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {k + 2} \paren {k + 1} x^k\) | Translation of Index Variable of Product | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 {\paren {1 + x}^3}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k\) |
$\blacksquare$
Proof 2
| \(\ds \frac 1 {\paren {1 + x} }\) | \(=\) | \(\ds \paren {1 + x}^{-3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-3}^{\underline k} } {k!} x^k\) | General Binomial Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-3} - j} } {k!} x^k\) | Definition of Falling Factorial | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 3} } } {\ds \prod_{j \mathop = 1}^k j} x^k\) | Definition of Factorial and simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-\paren {j + 2} } } {\ds \prod_{j \mathop = 1}^k j} x^k\) | Translation of Index Variable of Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k \paren {j + 2} } {\ds \prod_{j \mathop = 1}^k j} x^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k\) | simplification |
$\blacksquare$