Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 7

Table of contents
- a 7
  - Solution a
- b 7
  - Solution b

a

Solution

The binomial theorem gives

(1 + 0.01)^{9} = 1 + 9 (0.01) + 36 (0.01)^{2} + 84 (0.01)^{3} + 126 (0.01)^{4} + \dots = 1.093685272 \dots .

Thus, to three decimal places,

(1.01)^{9} ≃ 1.094.

The first three nonconstant terms already determine this rounding; the remaining positive terms total less than

1.3 \times 1 0^{- 6}

.

b

Solution

The initial term at

n = - 1

is three, and the common ratio is

1/3

. The convergent geometric series therefore sums to

n = - 1 \sum \infty (\frac{1}{3})^{n} = \frac{3}{1 - 1/3} = \frac{9}{2} .

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