Chapter 9: Sequences and Series Exercise 9.1

Given, a_n = 3a_n-1 + 2 and a₁ = 3

Then,

a₂ = 3a₁ + 2 = 3(3) + 2 = 11

a₃ = 3a₂ + 2 = 3(11) + 2 = 35

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

a₅ = 3a₄ + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

Given,

a_n = a_n-1/n and a₁ = -1

Then,

a₂ = a₁/2 = -1/2

a₃ = a₂/3 = -1/6

a₄ = a₃/4 = -1/24

a₅ = a₄/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

Given,

a₁ = a₂, a_n = a_n-1 – 1

Then,

a₃ = a₂ – 1 = 2 – 1 = 1

a₄ = a₃ – 1 = 1 – 1 = 0

a₅ = a₄ – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

Given,

1 = a₁ = a₂

a_n = a_{n – 1} + a_{n – 2}, n > 2

So,

a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Thus,