Question

Give the first 8 terms of the sequence. a1= -1, a2=2, a[n]=a[n-2](3-a[n-1])

Answer 1

Step-by-step explanation:

Give the first 8 terms of the sequence. a1= -1, a2=2, a[n]=a[n-2](3-a[n-1])

Given

first term a1 = -2

second term a2 = 2

We are to get the first 8 terms. Given the sequence

a[n]=a[n-2](3-a[n-1])

a[3]=a[3-2](3-a[3-1])

a[3]=a[1](3-a[2])

a[3]=-2(3-2)

a3 = -2

a[n]=a[n-2](3-a[n-1])

a[4]=a[4-2](3-a[4-1])

a[4]=a[2](3-a[3])

a[4]=2(3+2)

a4= 10

a[5]=a[n-2](3-a[n-1])

a[5]=a[5-2](3-a[5-1])

a[5]=a[3](3-a[4])

a[5]=-3(3-10)

a5 = -3(-7)

a5 = 21

a[6]=a[n-2](3-a[n-1])

a[6]=a[6-2](3-a[6-1])

a[6]=a[4](3-a[5])

a[6]= 10(3-21)

a6 = 10(-18)

a6 = -180

a[n]=a[n-2](3-a[n-1])

a[7]=a[7-2](3-a[7-1])

a[7]=a[5](3-a[6])

a[7]= 21(3+180)

a7 = 21(183)

a7 = 3,843

a[8]=a[n-2](3-a[n-1])

a[8]=a[8-2](3-a[8-1])

a[8]=a[6](3-a[7])

a[8]=-180(3-3843)

a8 = -180(-3840)

a8 = 691,200