Addition & Subtraction of Surd

Examples; Simplify the following:

Solution

Note;

Figure with figure and root with root; this rule simply means that you'll bring the figures together and take the root as one common factor. See below!

Practice these questions below;

Sequence

A Sequence is a succession of terms in such a way that they are related to one another according to a well defined rule. However, the rules may differ depending on the arrangement of such terms.

This means that the rule that works for a particular sequence may not necessarily work for another.

Note; The nth term of a sequence is denoted as T_n where n stands for number of terms.

Examples of sequence are;

6, 11, 16, 21, 26,…(they differ by + 5)

1, 3, 9, 27, 81,…(they differ by × 3)

1, ³/_2, 2, ⁵/₂, 3,…(they differ by + ½ )

Examples;

1. The nth term of a sequence is given by 3 × 2^{n -2}. Write down the first three terms of the sequence.

Solution

Step 1

Remember that above the nth term is denoted as T_n.

T_n = 3 × 2^{n -2}

T₁ = 3 × 2^{1 -2}

T₁ = 3 × 2^-1

T₁ = 3 × ½ (Any number raise to power of negative 1 must become inverse, it’s a law!)

T₁ = ³/₂

Step 2

T₂ = 3 × 2^{2 -2}

T₂ = 3 × 2⁰ (Note; Any number raise to power of zero must be 1)

T₂ = 3 × 1

T₂ = 3

Step 3

T₃ = 3 × 2^{3 -2}

T₃ = 3 × 2¹

T₃ = 3 × 2

T₃ = 6

Step 4

 The first three terms of the sequence are ^3/₂, 3, 6, …

 
2. If the nth term of a sequence is denoted by the formula: n(2^{n + 1}) – 3n, find the sum of the first four terms.

Solution

Step 1

T_n = n(2^{n + 1}) – 3n ----This the nth term formula

T₁ = 1(2^{1 + 1}) – 3(1)

T₁ = 1(2²) – 3

T₁ = 1(2 × 2) – 3

T₁ = 1(4) – 3

T₁ = 4 – 3

T₁ = 1

Step 2

T₂ = 2(2^{2 + 1}) – 3(2)

T₂ = 2(2³) – 6

T₂ = 2(2 × 2 × 2) – 6

T₂ = 2(8) – 6

T₂ = 16 – 6

T₂ = 10

Step 3

T₃ = 3(2^{3 + 1}) – 3(3)

T₃ = 3(2⁴) – 9

T₃ = 3(2 × 2 × 2 × 2) – 9

T₃ = 3(16) – 9

T₃ = 48 – 9

T₃ = 39

Step 4

T₄ = 4(2^{4 + 1}) – 3(4)

T₄ = 4(2⁵) – 12

T₄ = 4(2 × 2 × 2 × 2 × 2) – 12

T₄ = 4(32) – 12

T₄ = 128 – 12

T₄ = 116

 The sum of the first four terms will be; 1 + 10 + 39 + 116 = 166.

                                                                                                     

3. Given the nth term of a sequence as

     

Find the first four terms.

Solution

Step 1

 

T₁ = - ¼

Step 2

T₂ = 0

Step 3

T₃ = ¾

Step 4

  

T₄ = 2

 The first four terms of the sequence are; - ¼, 0, ¾, 2, …

Practice these questions below;

1. The nth term of a sequence is denoted by 3n(2ⁿ – 1). Find the sum of the first five terms.

2. A particular term of sequence is represented by the formula 3 × 2^{n + 2}. What is the sum of the 5^th and 6^th terms?

3. Find the difference between the 4^th and 11^th terms of the sequence whose nth term is

4. Given that the nth term of a sequence is denoted by the formula;

Write down the first six terms.

5. What is the product of the 9^th and 12^th terms of the sequence with the nth term equals 4n² – 2^{n – 1}.

Series

A Series is the addition of successive terms in such a way that they are related to another according to a well defined rule. In other words, series is the addition of a sequence.

Examples of series are;

2 + 5 + 8 + 11 + 14 + ……..

4 + 16 + 64 + 256 + 1024 + ………

-1 + 0 + 7 + 14 + 23 + 34 + ……..

Note

Just like sequence you will be given what you’ll know as nth term formula to find the terms of series and the formula always differ per questions.

Examples;

1. Find the series of the first six terms of 2ⁿ + 4n².

Solution

Note

2ⁿ + 4n² stands as a formula that you’ll use to solve for the first six terms mentioned in the question. Finding the first six terms means solving for T_1, T_2, T_3, T_4, T₅ and T₆ respectively_.

Step 1

T_n = 2ⁿ + 4n² ---The nth term formula.

T₁ = 2¹ + 4(1)²

T₁ = 2 + 4(1²)

T₁ = 2 + 4(1)

T₁ = 2 + 4

T₁ = 6

Step 2

T₂ = 2² + 4(2)²

T₂ = 4 + 4(2²)

T₂ = 4 + 4(2 × 2)

T₂ = 4 + 4(4)

T₂ = 4 + 4 × 4

T₂ = 4 + 16

T₂ = 20

Step 3

T₃ = 2³ + 4(3)²

T₃ = 8 + 4(3²)

T₃ = 8 + 4(3 × 3)

T₃ = 8 + 4(9)

T₃ = 8 + 36

T₃ = 44

Step 4

T₄ = 2⁴ + 4(4)²

T₄ = 16 + 4(4²)

T₄ = 16 + 4(4 × 4)

T₄ = 16 + 4(16)

T₄ = 16 + 64

T₄ = 80

Step 5

T₅ = 2⁵ + 4(5)²

T₅ = 32 + 4(5²)

T₅ = 32 + 4(5 × 5)

T₅ = 32 + 4(25)

T₅ = 32 + 100

T₅ = 132

Step 6

T₆ = 2⁶ + 4(6)²

T₆ = 64 + 4(6²)

T₆ = 64 + 4(6 × 6)

T₆ = 64 + 4(36)

T₆ = 64 + 144

T₆ = 208

 The series of the first six terms will be; 6 + 20 + 44 + 80 + 132 + 208 +…

2. Find the sum of the series n² + 5n up to the 4^th term.

Solution

Step 1

T_n = n² + 5n---Formula

T₁ = 1² + 5(1)

T₁ = 1 + 5

T₁ = 6

Step 2

T₂ = 2² + 5(2)

T₂ = 4 + 5 × 2

T₂ = 4 + 10

T₂ = 14

Step 3

T₃ = 3² + 5(3)

T₃ = 9 + 5 × 3

T₃ = 9 + 15

T₃ = 24

Step 4

T₄ = 4² + 5(4)

T₄ = 16 + 5 × 4

T₄ = 16 + 20

T₄ = 36

 The sum of the series will be; 6 + 14 + 24 + 36 = 80.

 

Practice these questions below;

1. What is the sum of the eight series of an nth term equals 2ⁿ + 5n?

2. Find the seven series of the term (n + 1)² – 2n.

3. Write out the series of the term 3ⁿ(n – 1).

4. Find the series of the first five terms of n(2 + 3n).

5. Find the sum of the series 5(3^{n - 1}) up to the 4^th term.