Exercise 47 Page 319

The given rule means that after the first term of the sequence, every term f(n) is the sum of the previous term f(n-1) and 7.

f(2) = 10, f(5) = 31, f(10) = 66

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We are asked to write the 2nd, 5th and 10th terms of a sequence, given a recursive rule. \begin{aligned} f(1)&=3 \\ f(n)&=f(n-1)+7, \text{ for } n\gt 1 \end{aligned} Let's use a table.

n	f(n)=f(n-1)+7	f(n)
1	f( 1)=3	3
2	f( 2)=f( 2-1)+7 ⇕ f(2)= f(1)+7	f(2)= 3+7 ⇕ f(2)= 10
3	f( 3)=f( 3-1)+7 ⇕ f(3)= f(2)+7	f(3)= 10+7 ⇕ f(3)= 17
4	f( 4)=f( 4-1)+7 ⇕ f(4)= f(3)+7	f(4)= 17+7 ⇕ f(4)= 24
5	f( 5)=f( 5-1)+7 ⇕ f(5)= f(4)+7	f(5)= 24+7 ⇕ f(5)= 31
6	f( 6)=f( 6-1)+7 ⇕ f(6)= f(5)+7	f(6)= 31+7 ⇕ f(6)= 38
7	f( 7)=f( 7-1)+7 ⇕ f(7)= f(6)+7	f(7)= 38+7 ⇕ f(7)= 45
8	f( 8)=f( 8-1)+7 ⇕ f(8)= f(7)+7	f(8)= 45+7 ⇕ f(8)= 52
9	f( 9)=f( 9-1)+7 ⇕ f(9)= f(8)+7	f(9)= 52+7 ⇕ f(9)= 59
10	f( 10)=f( 10-1)+7 ⇕ f(10)= f(9)+7	f(10)= 59+7 ⇕ f(10)= 66

Therefore, f(2) = 10, f(5) = 31 and f(10) = 66.

Subchapter links

Communicate Your Answer p.313

Monitoring Progress p.315-317

Exercises p.318-320

Exercise 47 Page 319

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