Exercise 51 Page 345

The given rule means that after the first term of the sequence, every term f(n) is the difference of the previous terms f(n-1) and f(n-2).

f(2) = 15 f(5) = - 15, f(10) = - 10

Practice makes perfect

We are asked to write the first 2^\text{nd}, 5^\text{th} and 10^\text{th} term of a sequence, given a recursive rule. f(1)&=10 f(2)&=15 f(n)&=f(n-1)-f(n-2), for n>2 Let's use a table.

n	f(n)=f(n-1)-f(n-2)	f(n)
1	f( 1)=10	10
2	f( 2)=15	15
3	f( 3)=f( 3-1)-f( 3-2) ⇕ f(3)= f(2)- f(1)	f(3)= 15- 10 ⇕ f(3)= 5
4	f( 4)=f( 4-1)-f( 4-2) ⇕ f(4)= f(3)- f(2)	f(4)= 5- 15 ⇕ f(4)= - 10
5	f( 5)=f( 5-1)-f( 5-2) ⇕ f(5)= f(4)- f(3)	f(5)= - 10- 5 ⇕ f(5)= - 15
6	f( 6)=f( 6-1)-f( 6-2) ⇕ f(6)= f(5)- f(4)	f(6)= - 15-( - 10) ⇕ f(6)= - 5
7	f( 7)=f( 7-1)-f( 7-2) ⇕ f(7)= f(6)- f(5)	f(7)= - 5-( - 15) ⇕ f(7)= 10
8	f( 8)=f( 8-1)-f( 8-2) ⇕ f(8)= f(7)- f(6)	f(8)= 10-( - 5) ⇕ f(8)= 15
9	f( 9)=f( 9-1)-f( 9-2) ⇕ f(9)= f(8)- f(7)	f(9)= 15- 10 ⇕ f(9)= 5
10	f( 10)=f( 10-1)-f( 10-2) ⇕ f(10)= f(9)- f(8)	f(10)= 5- 15 ⇕ f(10)= - 10

Therefore, f(2) = 15, f(5) = - 15 and f(10) = - 10.

Subchapter links

Communicate Your Answer p.339

Monitoring Progress p.341-343

Exercises p.344-346

Exercise 51 Page 345

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