Closed Form for Pentagonal Pyramidal Numbers

Theorem

The closed-form expression for the $n$th pentagonal pyramidal number is:

$Q_n = \dfrac {n^2 \paren {n + 1} } 2$

Proof

 $\ds Q_n$ $=$ $\ds \sum_{k \mathop = 1}^n P_k$ Definition of Pentagonal Pyramidal Number $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \dfrac {k \paren {3 k - 1} } 2$ Closed Form for Pentagonal Numbers $\ds$ $=$ $\ds \dfrac 1 2 \paren {3 \sum_{k \mathop = 1}^n k^2 - \sum_{k \mathop = 1}^n k}$ $\ds$ $=$ $\ds \dfrac 1 2 \paren {3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - \sum_{k \mathop = 1}^n k}$ Sum of Sequence of Squares $\ds$ $=$ $\ds \dfrac 1 2 \paren {3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - \dfrac {n \paren {n + 1} } 2}$ Closed Form for Triangular Numbers $\ds$ $=$ $\ds \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {2 n + 1} - n \paren {n + 1} } 2}$ $\ds$ $=$ $\ds \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {2 n} } 2}$ $\ds$ $=$ $\ds \frac {n^2 \paren {n + 1} } 2$

$\blacksquare$