Question

The table below shows three special sequences (A to C) of numbers.

Answer 1

Answer:

3.1 A) 49; 64; and 81

B) 343; 512; and 729

C) 28; 36; and 45

3.2 A) 1²; 2²; 3²...

B) 1³; 2³; 3³...

C) 1; 1+2; 1+2+3; ...

3.3 A) aₙ = n²

B) aₙ = n³

C) aₙ = n·(n + 1)/2

3.4 A) 10,000

B) 1,000,000

C) 5,050

Step-by-step explanation:

The given sequences are;

A 1; 4; 9; 16; 25; 36...

B 1; 8; 27; 64; 125; 216...

C 1; 3; 6; 10; 15; 21;...

3.1 A) The terms, aₙ, of sequence 'A' are given as follows;

aₙ = n², where, n = 1, 2, 3,...

The next three terms of sequence 'A' are therefore;

a₇ = 7³ = 49, a₈ = 8² = 64, and a₉ = 9² = 81

The next three terms of sequence 'A' are ; 49; 64; and 81

B) The terms, aₙ, of sequence 'B' are given as follows;

aₙ = n³, where, n = 1, 2, 3,...

The next three terms of sequence 'B' are therefore;

a₇ = 7³ = 343, a₈ = 8³ = 512, and a₉ = 9³ = 729

The next three terms of sequence 'B' are; 343; 512; and 729

C) The terms, aₙ, of sequence 'C' are given as follows;

aₙ = n·(n + 1)/2, where, n = 1, 2, 3,...

The next three terms of sequence 'A' are therefore;

a₇ = 7×(7 + 1)/2 = 28, a₈ = 8×(8 + 1)/2 = 36, and a₉ = 36 + 9×(9 + 1)/2 = 45

The next three terms of sequence 'A' are ; 28; 36; and 45

3.2 A) The pattern of the numbers in sequence 'A', consists of squaring (raising to the power of 2) each number in the sequence of natural numbers

B) The pattern of the numbers in sequence 'B', consists of raising to the power of 3 each number in the sequence of natural numbers

C) The pattern is the triangular number sequence and consists of finding half the number dots that are present in a n by (n + 1) rectangle, which is given by finding the sum of the natural numbers up to the given term in the sequence

3.3 A) The formula for the general term of sequence, A, is aₙ = n²

B) The formula for the general term of sequence, B, is aₙ = n³

C) The formula for the general term of sequence, C, is aₙ = n·(n + 1)/2

3.4 A) The 100th term of sequence 'A', a₁₀₀ = 100² = 10,000

B) The 100th term of sequence 'B', a₁₀₀ = 100³ = 1,000,000

C) The 100th term of sequence 'C', a₁₀₀ = 100 × (100 + 1)/2 = 5,050.