Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 7

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

a

Solution

There are

100

terms in this arithmetic series, starting at

4

and ending at

202

. Pairing the first and last terms, the second and penultimate terms, and so on, gives

n = 2 \sum 101 2 n = \frac{100}{2} (4 + 202) = 10300.

The summand in the PDF is

2 n

, not

2^{n}

.

b

Solution

This is a finite geometric series with first term and common ratio both equal to four. Therefore

n = 1 \sum 5 4^{n} = \frac{4 ( 4 ^{5} - 1 )}{4 - 1} = 1364.

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