Chapter 9 – Sequences and Series (Ex – 9.1)

Sequences and Series (Ex – 9.1)

Question 1.
a_n = n(n + 2)

Solution:
We haven a_n = n(n + 2)
substituting n = 1, 2, 3, 4, 5, we get
a₁ = 9(1 + 2) = 1 x 3 = 3
a₂ = 2(2 + ) = 2 x 4 = 8
a₃ = 3(3 + 2) = 3 x 5 = 15
a₄ = 4(4 + 2) = 4 x 6 = 24
a₅ = 5(5 + 2) = 5 x 7 = 35
∴ The first five terms are 3, 8, 15, 24, 35.

Question 2.
a_n = n/(n+1)

Solution:

Question 3.
a_n = 2ⁿ

Solution:

Question 4.
a_n = (2n−3)/6

Solution:

Question 5.
a_n = (- 1)^n-1 5ⁿ⁺¹

Solution:
We have, a_n = (- 1)^n-1 5ⁿ⁺¹
Substituting n = 1, 2, 3, 4, 5, we get
a₁ =(-1)^1-1 5¹⁺¹ = (-1)^° 5² = 25
a₂ =(-1)^2-1 5²⁺¹ = (-1)¹ 5³ = 125
a₃ =(-1)^3-1 5³⁺¹ = (-1)² 5⁴ = 625
a₄ =(-1)^4-1 5⁴⁺¹ = (-1)³ 5⁵ = -3125
a₅ =(-1)^5-1 5⁵⁺¹ = (-1)⁴ 5⁶ = 15625
∴ The first five terms are 25, – 125, 625, -3125, 15625.

Question 6.

Solution:
Substituting n = 1, 2, 3, 4, 5, we get

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

Question 7.
a_n = 4n – 3; a₁₇, a₂₄

Solution:
We have a_n = 4n – 3

Question 8.
a_n = n²/2n; a₇

Solution:
We have, a_n = n²/2n; a₇

Question 9.
a_n = (-1)^{n – 1} n³; a₉

Solution:
We have, a_n = (-1)^{n – 1} n³

Question 10.

Solution:
We have,

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Question 11.
a₁ = 3, a_n = 3a_n-1+2 for all n>1

Solution:
We have given a₁ = 3, a_n = 3a_n-1+2
⇒ a₁ = 3, a₂ = 3a₁ + 2 = 3.3 + 2 = 9 + 2 = 11,
a₃ = 3a₂ + 2 = 3.11 + 2 = 33 + 2 = 35,
a₄ = 3a₃ + 2 = 3.35 + 2 = 105 + 2 = 107,
a₅ = 3a₄ + 2 = 3.107 + 2 = 321 + 2 = 323,
Hence, the first five terms of the sequence are 3, 11, 35, 107, 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..

Question 12.

Solution:
We have given

Hence the first five terms of the given sequence are – 1, -1/2, -1/6, -1/24, -1/120.
The corresponding series is

Question 13.
a₁ = a₂ = 2, a_n = a_n-1 – 1, n >2

Solution:
We have given a₁ = a₂ = 2, a_n = a_n-1 – 1, n >2
a₁ = 2, a₂ = 2, a₃= a₂ – 1 = 2 – 1 = 1,
a₄ = a₃ – 1 = 1 – 1 = 0 and a₅ = a₄ – 1 = 0 – 1 = -1
Hence the first five terms of the sequence are 2, 2, 1, 0, -1
The corresponding series is
2 + 2 + 1 + 0 + (-1) + ……

Question 14.

Solution:
We have,