The diagram above shows the pentagon growing in a systematic way to produce the pentagonal numbers. The equations below describe this growth:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{P_2}& =& P_1+4\\ \multicolumn{1}{c}{}& =& 1+4\\ \multicolumn{1}{c}{}& =& 1+3\times 1+1\\ \multicolumn{1}{c}{P_3}& =& P_2+7\\ \multicolumn{1}{c}{}& =& 1+4+7\\ \multicolumn{1}{c}{}& =& 1+(3\times 1+1)+(3\times 2+1)\\ \multicolumn{1}{c}{P_4}& =& P_3+10\\ \multicolumn{1}{c}{}& =& 1+(3\times 1+1)+(3\times 2+1)+(3\times 3+1)\\ \multicolumn{1}{c}{}\end{array}}$

Can you find a general rule for $P_n$?

By writing a formula for the nth triangular number $T_n$, show that all pentagonal numbers are one third of a triangular number and prove that the triangular numbers involved are all of the form $T_{3n-1}$