Exercise 8 Page 447

The given rule means that, after the first term of the sequence, every term a_n is the difference between the square of the previous term a_(n-1) and 10.

a_1=1, a_2=- 9, a_3=71, a_4=5031, a_5=25 310 951, a_6=640 644 240 524 391

Practice makes perfect

We are asked to write the first 6 terms of a sequence, given a recursive rule. a_1&=1 a_n&=(a_(n-1))^2-10 To do so, we will use a table.

n	a_n=(a_(n-1))^2-10	(a_(n-1))^2-10	a_n
1	a_1=1	-	1
2	a_2=(a_(2-1))^2-10	( a_1)^2-10= 1^2-10	- 9
3	a_3=(a_(3-1))^2-10	( a_2)^2-10=( - 9)^2-10	71
4	a_4=(a_(4-1))^2-10	( a_3)^2-10= 71^2-10	5031
5	a_5=(a_(5-1))^2-10	( a_4)^2-10= 5031^2-10	25 310 951
6	a_6=(a_(6-1))^2-10	( a_5)^2-10= 25 310 951^2-10	640 644 240 524 391

Therefore, the first 6 terms of the sequence are 1, - 9, 71, 5031, 25 310 951 and 640 644 240 524 391.

Subchapter links

Communicate Your Answer p.441

Monitoring Progress p.442-446

Exercises p.447-450

Exercise 8 Page 447

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