Cambridge International AS and A Level Mathematics Pure Mathematics 1

Geometric progressions

P1^

3

Now multiply it by the common ratio, 2:

2 S = 2 + 4 + 8 + 16 + ... + 264. ^2

Then subtract ^1 from ^2

^2  2 S^ =^2 +^4 +^8 +^16 +^ ... +^263 +^264

^1 S^ =^1 +^2 +^4 +^8 +^ ... +^263

subtracting: S = –1 + 0 + 0 + 0 + ... + 264.

The total number of rice grains requested was therefore 2^64 − 1 (which is about

1.85 × 1019 ).

●?^ How many tonnes of rice is this, and how many tonnes would you expect there

to be in China at any time?

(One hundred grains of rice weigh about 2 grammes. The world annual

production of all cereals is about 1.8 × 109 tonnes.)

Note

The method shown above can be used to sum any geometric progression.

ExamPlE 3.8 Find the value of 0.2 + 1 + 5 + ... + 390 625.

SOlUTION

This is a geometric progression with common ratio 5.

Let S = 0.2 + 1 + 5 + ... + 390 625. ^1

Multiplying by the common ratio, 5, gives:

5 S = 1 + 5 + 25 + ... + 390 625 + 1 953 125. ^2

Subtracting ^1 from ^2 :

5 S = 1 + 5 + 25 + ... + 390 625 + 1 953 125

S = 0.2 + 1 + 5 + 25 + ... + 390 625

4 S = −0.2 + 0 + ... + 0 + 1 953 125

This gives 4 S = 1 953 124.8

⇒ S = 488 281.2.