Exercise 7 Page 447

The given rule means that, after the first term of the sequence, every term a_n is the sum of the square of the previous term a_n-1 and 1.

a_1=2, a_2=5, a_3=26, a_4=677, a_5=458 330, a_6=210 066 388 901

Practice makes perfect

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

n	a_n=(a_(n-1))^2+1	(a_(n-1))^2+1	a_n
1	a_1=2	-	2
2	a_2=(a_(2-1))^2+1	( a_1)^2+1= 2^2+1	5
3	a_3=(a_(3-1))^2+1	( a_2)^2+1= 5^2+1	26
4	a_4=(a_(4-1))^2+1	( a_3)^2+1= 26^2+1	677
5	a_5=(a_(5-1))^2+1	( a_4)^2+1= 677^2+1	458 330
6	a_6=(a_(6-1))^2+1	( a_5)^2+1= 458 330^2+1	210 066 388 901

Therefore, the first 6 terms of the sequence are 2, 5, 26, 677, 458 330 and 210 066 388 901.

Communicate Your Answer p.441

Monitoring Progress p.442-446

Exercises p.447-450

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