Cambridge International AS and A Level Mathematics Pure Mathematics 1

Sequences and series

P1^

3

ExamPlE 3.3 Find the value of 8 + 6 + 4 + ... + (−32).

SOlUTION

This is an arithmetic progression, with common difference −2. The number of

terms, n, may be calculated using

n=+la–d 1

n=+––^32 – 28 1

= 21.

The sum S of the progression is then found as follows.

S = 8 + 6 + ... − 30 − 32

S = −32 – 30 − ... + 6 + 8

2 S = − 24 − 24 − ... − 24 − 24

Since there are 21 terms, this gives 2S = − 24 × 21, so S = − 12 × 21 = −252.

Generalising this method by writing the series in the conventional notation gives:

Sn = [a] + [a + d] + ... + [a + (n − 2)d] + [a + (n − 1)d]

Sn = [a + (n − 1)d] + [a + (n − 2)d] + ... + [a + d] + [a]

2 Sn = [2a + (n − 1)d] + [2a + (n − 1)d] + ... + [2a + (n − 1)d] + [2a + (n − 1)d]

Since there are n terms, it follows that

Snn=+^1 [ an()− d]

2

21

This result may also be written as

Snn=+^12 ()al.

ExamPlE 3.4 Find the sum of the first 100 terms of the progression

11 ,,^141112 ,,^34 .

SOlUTION

In this arithmetic progression

a = 1, d = 14 and n = 100.

Using Snn=+^12 [ 21 an(– )d], you have

Sn=× 21 100 () 29 +× (^914)
= 1337^12.