Generating Function for Triangular Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T_n$ denote the $n$th triangular number.


Then the generating function for $\sequence {T_n}$ is given as:

$\ds \map G z = \frac z {\paren {1 - z}^3}$


Corollary

Let $\sequence {b_n}$ be the sequence defined as:

$\forall n \in \N_{> 0}: b_n = \dfrac {\paren {n + 1} \paren {n + 2} } 2$

That is:

$\sequence {b_n}_{n \mathop \ge 0} = 1, 3, 6, 10, \ldots, \dbinom {n + 2} 2, \ldots$


Then the generating function for $\sequence {b_n}$ is given as:

$H \paren z = \dfrac 1 {\paren {1 - z}^3}$


Proof

\(\ds \frac z {\paren {1 - z}^3}\) \(=\) \(\ds z \paren {1 - z}^{-3}\) Exponent Combination Laws for Negative Power
\(\ds \) \(=\) \(\ds z \sum_{n \mathop = 0}^\infty \dbinom {-3} n \paren {-z}^n\) General Binomial Theorem
\(\ds \) \(=\) \(\ds z \sum_{n \mathop = 0}^\infty \paren {-1}^n \dbinom {n + 2} n \paren {-z}^n\) Negated Upper Index of Binomial Coefficient
\(\ds \) \(=\) \(\ds z \sum_{n \mathop = 0}^\infty \dbinom {n + 2} n z^n\)
\(\ds \) \(=\) \(\ds z \sum_{n \mathop = 0}^\infty \dbinom {n + 2} 2 z^n\) Symmetry Rule for Binomial Coefficients
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dbinom {n + 2} 2 z^{n + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \dbinom {n + 1} 2 z^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty T_n z^n\) Corollary to Binomial Coefficient with Two
\(\ds \) \(=\) \(\ds 0 z^0 + \sum_{n \mathop = 1}^\infty T_n z^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty T_n z^n\) $T_0 = 0$ by Definition 1 of Triangular Number
\(\ds \) \(=\) \(\ds \map G z\) Definition of Generating Function

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Generating_Function_for_Triangular_Numbers&oldid=682338"