tice that 1(2n1) 2 n 3 (2n3), and so on. Adding the right sides,
we get n sums of 2n, or 2n^2. So 2S 2 n^2 , and the result follows.
The second way is so easy that third graders to whom I’ve presented talks
understand the idea. It requires looking at these sums on a checkerboard.
The number 1 is represented by the square in the upper-left-hand corner of
the checkerboard. The number 3 is represented by all the squares in the sec-
ond row or column that share a vertex with the upper-left-hand corner square.
Together, 1 3 makes up the square in the upper left hand corner that is two
checkerboard squares on a side. The number 5 is represented by all the
squares in the third row or column that share a vertex with a square used in the
representation of 3. Together, 1 3 5 makes up the square in the upper-left-
hand corner that is three checkerboard squares on a side. And so on.
You can also use the principle of mathematical induction.
The line
1 12establishes the proposition (the sum of the first n odd numbers is n^2 ) for
n1. If we assume the proposition is true for the integer n, all we need do
is to show that the proposition is true for n1. This proposition would be
that the sum of the first n1 odd numbers is (n1)^2. Written formally,
we need to establish that, under the assumption1 3 5 (2n1)n^2 (the formula is valid for the
integer n)we can proceed to prove
1 3 5 (2(n1)1)(n1)^2 (the formula is valid for
the integer n1)
The basic facts of algebraic and arithmetic manipulation can be deduced
from the Peano axioms, but to do so is somewhat technical, and so for the
remainder of this proof we’ll just assume the usual laws of arithmetic
and algebra, such as abba.
Simplifying the expression in parentheses on the left side of the equa-
tion yields
1 3 5 (2n1)(n1)^2Continuing, we obtain
1 3 5 (2n1)[1 3 5 (2n1)](2n1)
n^2 (2n1) (this substitution is our assumption)
(n1)^2 (basic algebra)
Even Logic Has Limits 121