2 0. Background
(u2 + d)+(uN−1 − d) = u2 + uN−1 = u1 + uN . Continuing in this way, we find that
uk + uN+1−k = u1 + uN for all k. Hence the right hand side above is N(u1 + uN ),
so we have the result.
A sequence {un} is said to be a geometric progression if un+1/un is the same
for all n. That means that un =
arn
for some a and r.
Theorem 0.2. The sum of a geometric progression is
(0.2) 1 + r +
r2
+ · · · +
rN−1
=
⎧
⎪1
⎨
⎪
⎩
−
rN
1 − r
(r = 1),
N (r = 1).
If the first term to be summed is a power of r, we can simply factor out that
amount. Thus for example,
M+N−1
n=M
rn
=
rM − rM+N
1 − r
if r = 1. If |r| 1, then rN → 0 as N → ∞, so
(0.3)
∞
n=0
rn
=
1
1 − r
.
Proof. If r = 1, then each of the N terms on the left is 1, so the sum is N. Suppose
that r = 1, and let S denote the sum. Then
rS = r +
r2
+
r3
+ · · · +
rN
.
On subtracting this from S we see that most terms cancel, leaving
S − rS = 1 −
rN
.
The stated formula now follows on dividing both sides by 1 − r.
If p(x) = ax2 + bx + c is a quadratic polynomial, then
4ap(x) =
4a2x2
+ 4abx + 4ac = (2ax +
b)2
+ 4ac −
b2
.
This manipulation is called completing the square. If p(x) = 0, then (2ax + b)2 = d
where d = b2 − 4ac is the discriminant of the polynomial. If d 0, then the
equation p(x) = 0 has two distinct real roots, namely
r1 =
−b +
√
d
2a
, r2 =
−b −
√
d
2a
.
If d = 0, then p(x) has a double root, r1 = r2 = −b/(2a). If d 0, then p(x) has
no real root, but it has two complex roots,
−b ± i
√
−d
2a
.
In all three cases the sum of the roots is −b/a, the product of the roots is c/a, and
the polynomial factors, p(x) = a(x − r1)(x − r2).
