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.

Word Problems Leading To Linear Inequalities

Words problems leading to inequalities are handled in a similar fashion like the linear equations except that the equality sign (=) changes to inequality sign.

Examples;

1. One-fifth of a number added to two-third of the same number is greater than 26. Find the range of values of the number.

Solution

Step 1

Let x represent the unknown number

The mathematical interpretation of the question will be;

^x/₅ + ^2x/₃ > 26

Step 2

Multiply through by 15 which is the L.C.M of 5 and 3;

3x + 10x > 390

13x > 390

Step 3

Divide both sides by 13;

x > ³⁹⁰/₁₃

x > 30.

Therefore, the range of values of x will be numbers greater than 30, which are 31, 32, 33, 34, 35 …

2. Ade bought x biros at #8 each and (x + 4) rulers at #20 each. He spent less than #200. Form an inequality for the statement and solve it.

Solution

Step 1

Interpret the statement;

The biros cost #8x …….(1)

The rulers cost #20(x +4) …….(2)

Step 2

Expand the second statement and you will have;

(#20x + #80)

Step 3

Combining (1) and (2) gives the total cost of both biros and rulers;

#(8x + 20x + 80) which is less than (<) #200.

Step 4

Forming an inequality equation below, you’ll have;

8x + 20x + 80 < 200

Step 5

Re-arrange and collect like terms;

Hence 28x < 200 – 80 (This is the inequality statement)

28x < 120

Step 6

Divide both sides by 28;

x < ¹²⁰/₂₈

Therefore, x < 4.29.

Practice these questions below;

1. The sum of twice a number and 5 is less than the sum of one-third of the number and 6.

2. Two-thirds of a certain number is greater than the sum of the number and 6.

3. The sum of twice a number and 15 is less than thrice the same number minus 9.

4. A cyclist travels xkm in 4 hours, then (x + 60)km in 7 hours. Its average speed does not exceed 150km/h.

 
5. A boy bought x mangoes at #5 and 3x oranges at #6. He collected some balance from #30.