A PRIMER OF ANALYTIC NUMBER THEORY: From Pythagoras to ...

8 1. Sums and Differences
Figure 1.6. The pyramidal numbers P1 = 1, P2 = 5, P3 = 14, P4 = 30,....
and therefore,
Tn = n(n + 1)(n + 2)/6. (1.5)
Exercise 1.1.4. Use mathematical induction to give another proof of
Eq. (1.5), with Tn defined by Eq. (1.4).
The pyramidal numbers, Pn, give the number of objects in a pyramid
with a square base, as in Figure 1.6. The kth layer of the pyramid is a square
with sk = k 2 objects in it; so, by definition,
Pn = 1 2 + 2 2 + 3 2 +···+n 2 =
n�
k 2 .
Since we know a relationship between square numbers and triangular numbers,
we can get a formula for Pn in terms of the formula for Tn, as follows.
From Eq. (1.2) we have tk + tk−1 = k2 for every k. This even works for k = 1
if we define t0 = 0, which makes sense. So,
n�
Pn = k 2 n�
= {tk + tk−1}
=
k=1
n�
k=1
tk +
k=1
k=1
n�
tk−1 = Tn + Tn−1.
k=1
According to Eq. (1.5) this is just
Pn = n(n + 1)(n + 2)/6 + (n − 1)n(n + 1)/6
= n(n + 1)(2n + 1)/6.