Theorem 25.2 [On Arithmetic Sequences]

Suppose c_k+1 = c_k+a for k = 1...,n-1. Then for k = 1...,n, we have  c_k = c₁+(k-1)a

Proof. We want to show by mathematical induction that c_k = c₁+(k-1)a. This statement is true for k = 1. Moreover, if c_k = c₁+(k-1)a then c_k+1 = c_k+a = c₁+(k-1)a+a = c₁+ka. Thus if the assertion is true for k, it is also true for k+1. Thus the principle of mathematical induction implies the equality c_k = c₁+(k-1)a for each whole number k = 1,2,...,m.

Theorem 25.3 [On Arithmetic Sums]

Suppose  for k = 1...,n, we have  c_k = c₁+(k-1)a. Suppose

Sm = m å j = 1  cj = c1+c2+¼+cm

whenever 1 £ m £ n. Further for m = 1...,n (i)

Sm = m å j = 1  cj = mc1+ 1 2 m(m-1)a

and also (ii)

Sm = m å j = 1  cj = 1 2 k(c1+cm)

Note (ii) says that 2S_m is m times the sum of the first and last terms. The following proofs are optional readings in the first instance, but it mastering no matter how slowly will give you a taste hopefully not too bitter of higher mathematics. This author as a mathematical novice took two or three days, or even a week, to read and reread a proof until it was completely understood. Remember what is hard today may be easier tomorrow.

Proof of (i) . The equality c_k = c₁+(k-1)a just proven with k = m+1 implies c_m+1 = c₁+(m+1-1)a = c₁+ma.

We will prove

k å j = 1  cj = kc1+ 1 2 k(k-1)a

by induction on k ³ 1.

If k = 1 then S_m = å_{j = 1}^m c_j = c₁ and

kc1+ 1 2 k(k-1)a = 1c1+ 1 2 1(0)a = c1

Therefore the assertion is true for k = 1.

Now we want to show if the assertion

k å j = 1  cj = kc1+ 1 2 k(k-1)a

is true for k = m ³ 1 then it must be true when k = m+1. To this end, suppose

m å j = 1  cj = mc1+ 1 2 m(m-1)a

This and the equality

m+1 å j = 1  cj = cm+1+ m å j = 1  cj

imply

m+1 å j = 1  cj = cm+1+ é ê ë mc1+ 1 2 m(m-1)a ù ú û = [c1+ma]+ é ê ë mc1+ 1 2 m(m-1)a ù ú û = c1+mc1+ 1 2 m(m-1)a+ma = (1+m)c1+[ 1 2 m(m-1)+m]a = (1+m)c1+ 1 2 [m(m-1)+2m]a = (m+1)c1+ 1 2 [m(m-1+2)]a = (m+1)c1+ 1 2 [m(m+1)]a = (m+1)c1+ 1 2 [m(m+1)]a

This says that

k å j = 1  cj = kc1+ 1 2 k(k-1)a

when k = m+1. The principle of mathematical induction now applies. It gives the desired conclusion.

Proof of (ii) . We wish to show

k å j = 1  cj = 1 2 k(c1+ck)

Observe

1 2 k(c1+ck) = 1 2 k(c1+[c1+(k-1)a]) = kc1+ 1 2 k(k-1)a

Therefore the assertion

k å j = 1  cj = 1 2 k(c1+ck)

follows from

k å j = 1  cj = kc1+ 1 2 k(k-1)a

But the latter is true by the just-proven assertion (b). Thus assertion (c) must be true as well.

Chapter 25, Sections:  [Geometric Sums] [ 25 Arithmetic Sums ] [ 25 Factorial Definition ] [ 25 Product Notation ] [ 25 Notation for Sums ]

Next:  [Recursive Definition of Factorial Function n!]
         or Chapter 26, Proofs and Logic-What's Next