Exercise 1.2.48* (Pentagonal numbers)

Question

The integers $1,3,6,10,\ldots,n(n+1)/2,\ldots$ are called the triangular numbers because they are the numbers of dots needed to make successive triangular arrays of dots.For example, the number $10$ can be perceived as the number of acrobats in a human triangle, $4$ in a row at the bottom, $3$ at the next level, then $2$, then $1$ at the top. The square numbers are $1,4,9,\ldots,n^2,\ldots$. The pentagonal numbers, $1,5,12,22,\ldots,(3n^2-n)/2,\ldots$, can be seen in a geometric array in the following way. Start with $n$ equally spaced dots $P_1,P_2,\ldots,P_n$ on a straight line in a plane, with distance $1$ between consecutive dots. Using $P_1P_2$ as a base side, draw a regular pentagon in the plane. Similarly, draw $n-2$ additional regular pentagons on the base sides $P_1P_3,P_1P_4,\ldots,P_1P_n$. Mark dots at each vertex and at unit intervals along the sides of these pentagons. Prove that the total number of dots in the array is $(3n^2 -n)/2$. In general, if regular $k$-gons are constructed on the sides $P_1P_2,P_1P_3,\ldots,P_1P_N$, with dots marked again at unit intervals, prove that the total number of dots is $1+kn(n-1)/2 -(n-1)^2$. This is the $n$th $k$-gonal number.

Richard Ganaye · Accepted Answer

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See figures on Wikipedia:
\begin{verbatim}
https://en.wikipedia.org/wiki/Pentagonal_number
\end{verbatim}
\begin{proof} 
Write $p_n$ the $n$th pentagonal number (by convention $p_0 = 0$). The value $p_n$ is obtain by adding three sides containing $n$ marked dots. We must subtract the $2$ summits in common, so, for all integers $n\geq 1$,
$$p_n = p_{n-1} + 3n-2.$$
(For $n = 1$, we obtain $p_1 = p_0 + 1 = 1$.)
From 
\begin{align*}
p_n &= p_{n-1} + 3n-2,\
p_{n+1} &= p_{n} + 3n + 1,
\end{align*}
we obtain by subtraction $p_{n+1}- p_n = p_n - p_{n-1} + 3$, i.e. $p_{n+1} = 2p_n - p_{n-1} + 3$ ($n\in \N^*$). Thus, for all $n \in \N$,
$$p_{n+2} = 2 p_{n+1}  -p_n + 3.$$
We can rewrite this relation using matrices: for all $n \in \N$
$$
\begin{pmatrix} p_{n+1} \p_{n+2} \end {pmatrix} = 
\begin{pmatrix}   0 & 1\ -1 & 2 \end{pmatrix}
\begin{pmatrix} p_{n} \p_{n+1} \end {pmatrix} 
 +
 \begin{pmatrix} 0 \ 3 \end {pmatrix},
$$
i.e.
$$P_{n+1} = A P_n + B,$$
where 
$$P_n = \begin{pmatrix} p_{n} \p_{n+1} \end {pmatrix},
\qquad A = \begin{pmatrix}   0 & 1\ -1 & 2 \end{pmatrix},
\qquad B =  \begin{pmatrix} 0 \ 3 \end {pmatrix},
\qquad P_0 = \begin{pmatrix} 0 \ 1 \end {pmatrix}.
$$
By an immediate induction, we obtain
$$P_n = A^n P_0 + (A^{n-1} + A^{n-2} + \cdots + A+I) B.$$
Indeed, if $P_n = A^n P_0 + (A^{n-1} + A^{n-2} + \cdots + A+I) B$, then $$P_{n+1} = AP_n+B = A(A^n P_0 + (A^{n-1} + A^{n-2} + \cdots + A+I) B) + B = A^{n+1} P_0 + (A^{n} + A^{n-1} + \cdots + A+I) B.$$
We compute $A^n$ by standard methods.

Write $A = {\cal M}_{\cal B}(f)$, where ${\cal B} = (e_1,e_2)$ is the canonical basis of $\C^2$.
Then $\ker(f) = \langle e'_1 angle$, where $e'_1 = (1,1)$. Take for $e'_2$ any non collinear vector, for instance $e'_2 = (0,1)$. Then ${\cal B}' =(e'_1,e'_2)$ is a basis, and the transfer matrix is $P = \begin{pmatrix}   1 & 0\ 1& 1 \end{pmatrix}$. Then 
$$A' = {\cal M}_{\cal B'}(f) = P^{-1} A P =  \begin{pmatrix}   1 & 1\ 0& 1 \end{pmatrix}.
$$
Since $A' = I + N$, where $N^2 = 0$, the binomial formula applied to $I,N$ (such that $IN = NI$), gives $A'^n = I +n N$, so
$$A'^n = \begin{pmatrix}   1 & n\ 0 &  1 \end{pmatrix}.$$
Therefore
$$A^n = P A'^n P^{-1} = \begin{pmatrix}   1 -n & n\ -n &  1+n \end{pmatrix}.$$
Then 
\begin{align*} 
C_n &= A^{n-1}+A^{n-1} + \cdots + A + I \
&=  \begin{pmatrix}   \sum_{k=0}^{n-1} (1-k) &  \sum_{k=0}^{n-1} k\ -\sum_{k=0}^{n-1} k &  \sum_{k=0}^{n-1} (1-k) \end{pmatrix}\
&=   \begin{pmatrix}   n - \frac{n(n-1}{2} & \frac{n(n-1}{2}\ - \frac{n(n-1}{2} & n +  \frac{n(n-1}{2}\end{pmatrix},
\end{align*}
so $$C_n B = \begin{pmatrix} \frac{3n(n-1)}{2} \ * \end {pmatrix}.$$
This gives
$$
P_n = \begin{pmatrix} p_{n} \ * \end {pmatrix} = \begin{pmatrix}   1 -n & n\ -n &  1+n \end{pmatrix} \begin{pmatrix} 0 \ 1 \end {pmatrix} +  \begin{pmatrix} \frac{3n(n-1)}{2} \ * \end {pmatrix},
$$
so $p_n = n +\frac{3n(n-1)}{2} = \frac{3n^2-n}{2}$.

\bigskip The general case is almost identical. Here $p_n$ is the $n$th $k$-gonal number.

The induction relations are
$$p_n = p_{n-1} + rn - (r-1),\qquad  	ext{where } r = k-2,$$
and
$$p_{n+1} = 2p_n - p_{n-1} + r.$$
The same computing, with $B = \begin{pmatrix} 0 \ r \end {pmatrix}$, gives
$$
P_n = \begin{pmatrix} p_{n} \ * \end {pmatrix} = \begin{pmatrix}   1 -n & n\ -n &  1+n \end{pmatrix} \begin{pmatrix} 0 \ 1 \end {pmatrix} +  \begin{pmatrix} r \frac{n(n-1)}{2} \ * \end {pmatrix},
$$
thus
$$p_n = n + (k-2) \frac{n(n-1)}{2} = 1 + k \frac{n(n-1)}{2} - (n-1)^2.$$
\end{proof}

Exercise 1.2.48* (Pentagonal numbers)

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Exercise 1.2.48* (Pentagonal numbers)

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