Sure! Let's examine the given sequence:
[tex]\[ 0, \quad 1, \quad 8, \quad 27, \quad 64, \quad \ldots \][/tex]
First, we identify the position of each term in the sequence:
- The first term is [tex]\(0\)[/tex] (position [tex]\(n=0\)[/tex])
- The second term is [tex]\(1\)[/tex] (position [tex]\(n=1\)[/tex])
- The third term is [tex]\(8\)[/tex] (position [tex]\(n=2\)[/tex])
- The fourth term is [tex]\(27\)[/tex] (position [tex]\(n=3\)[/tex])
- The fifth term is [tex]\(64\)[/tex] (position [tex]\(n=4\)[/tex])
- and so on.
Next, we observe the relationship between the position [tex]\(n\)[/tex] and the value in the sequence.
For [tex]\( n = 0 \)[/tex]:
[tex]\[ 0 = 0^3 \][/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ 1 = 1^3 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ 27 = 3^3 \][/tex]
For [tex]\( n = 4 \)[/tex]:
[tex]\[ 64 = 4^3 \][/tex]
We notice that each term in the sequence is the cube of its position [tex]\(n\)[/tex].
So, we can express the [tex]\( n^{\text{th}} \)[/tex] term of the sequence as:
[tex]\[ a_n = n^3 \][/tex]
Therefore, the expression for the [tex]\( n^{\text{th}} \)[/tex] term of the sequence [tex]\( 0, \quad 1, \quad 8, \quad 27, \quad 64, \quad \ldots \)[/tex] is:
[tex]\[ a_n = n^3 \][/tex]
