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).