Answer :
Certainly! Let's analyze the sequence given to identify the pattern and derive an expression for the [tex]\( n \)[/tex]-th term.
The start of the pattern given is:
[tex]\[ 1 \rightarrow \frac{1}{4} \rightarrow \frac{1}{9} \rightarrow \frac{1}{16} \rightarrow \frac{1}{25} \][/tex]
We can observe the following:
1. The first term is [tex]\( 1 \)[/tex].
2. The second term is [tex]\( \frac{1}{4} \)[/tex].
3. The third term is [tex]\( \frac{1}{9} \)[/tex].
4. The fourth term is [tex]\( \frac{1}{16} \)[/tex].
5. The fifth term is [tex]\( \frac{1}{25} \)[/tex].
Let's denote each term in the sequence by [tex]\( a_n \)[/tex] where [tex]\( n \)[/tex] is the position of the term in the sequence (i.e., [tex]\( n=1 \)[/tex] for the first term, [tex]\( n=2 \)[/tex] for the second term, etc.).
### Step-by-step Analysis:
1. First term ([tex]\( n = 1 \)[/tex]):
[tex]\[ a_1 = 1 = \frac{1}{1^2} \][/tex]
2. Second term ([tex]\( n = 2 \)[/tex]):
[tex]\[ a_2 = \frac{1}{4} = \frac{1}{2^2} \][/tex]
3. Third term ([tex]\( n = 3 \)[/tex]):
[tex]\[ a_3 = \frac{1}{9} = \frac{1}{3^2} \][/tex]
4. Fourth term ([tex]\( n = 4 \)[/tex]):
[tex]\[ a_4 = \frac{1}{16} = \frac{1}{4^2} \][/tex]
5. Fifth term ([tex]\( n = 5 \)[/tex]):
[tex]\[ a_5 = \frac{1}{25} = \frac{1}{5^2} \][/tex]
From this analysis, we notice that each term can be expressed as the reciprocal of the square of its position in the sequence.
### General Formula:
From the observations, it is evident that the pattern for the [tex]\( n \)[/tex]-th term follows the formula:
[tex]\[ a_n = \frac{1}{n^2} \][/tex]
Hence, we can write down the expression for the [tex]\( n \)[/tex]-th term in this pattern as:
[tex]\[ a_n = \frac{1}{n^2} \][/tex]
So, the [tex]\( n \)[/tex]-th term of the pattern is:
[tex]\[ \boxed{\frac{1}{n^2}} \][/tex]
The start of the pattern given is:
[tex]\[ 1 \rightarrow \frac{1}{4} \rightarrow \frac{1}{9} \rightarrow \frac{1}{16} \rightarrow \frac{1}{25} \][/tex]
We can observe the following:
1. The first term is [tex]\( 1 \)[/tex].
2. The second term is [tex]\( \frac{1}{4} \)[/tex].
3. The third term is [tex]\( \frac{1}{9} \)[/tex].
4. The fourth term is [tex]\( \frac{1}{16} \)[/tex].
5. The fifth term is [tex]\( \frac{1}{25} \)[/tex].
Let's denote each term in the sequence by [tex]\( a_n \)[/tex] where [tex]\( n \)[/tex] is the position of the term in the sequence (i.e., [tex]\( n=1 \)[/tex] for the first term, [tex]\( n=2 \)[/tex] for the second term, etc.).
### Step-by-step Analysis:
1. First term ([tex]\( n = 1 \)[/tex]):
[tex]\[ a_1 = 1 = \frac{1}{1^2} \][/tex]
2. Second term ([tex]\( n = 2 \)[/tex]):
[tex]\[ a_2 = \frac{1}{4} = \frac{1}{2^2} \][/tex]
3. Third term ([tex]\( n = 3 \)[/tex]):
[tex]\[ a_3 = \frac{1}{9} = \frac{1}{3^2} \][/tex]
4. Fourth term ([tex]\( n = 4 \)[/tex]):
[tex]\[ a_4 = \frac{1}{16} = \frac{1}{4^2} \][/tex]
5. Fifth term ([tex]\( n = 5 \)[/tex]):
[tex]\[ a_5 = \frac{1}{25} = \frac{1}{5^2} \][/tex]
From this analysis, we notice that each term can be expressed as the reciprocal of the square of its position in the sequence.
### General Formula:
From the observations, it is evident that the pattern for the [tex]\( n \)[/tex]-th term follows the formula:
[tex]\[ a_n = \frac{1}{n^2} \][/tex]
Hence, we can write down the expression for the [tex]\( n \)[/tex]-th term in this pattern as:
[tex]\[ a_n = \frac{1}{n^2} \][/tex]
So, the [tex]\( n \)[/tex]-th term of the pattern is:
[tex]\[ \boxed{\frac{1}{n^2}} \][/tex]
