Answer :
To express each of these series in sigma (summation) notation, we need to identify the general formula for the nth term of each series and then formulate the summation accordingly.
### (a) [tex]\(1 + 3 + 5 + \cdots + 99\)[/tex]:
This is an arithmetic series with the first term [tex]\(a = 1\)[/tex] and a common difference [tex]\(d = 2\)[/tex].
The nth term of an arithmetic series is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
Substituting the values, we get:
[tex]\[ a_n = 1 + (n-1) \cdot 2 = 1 + 2n - 2 = 2n - 1 \][/tex]
So, the series can be represented as:
[tex]\[ \sum_{n=1}^{50} (2n - 1) \][/tex]
where [tex]\(n\)[/tex] runs from 1 to 50.
### (b) [tex]\(2 + 4 + 6 + \cdots + 100\)[/tex]:
This is also an arithmetic series with the first term [tex]\(a = 2\)[/tex] and a common difference [tex]\(d = 2\)[/tex].
The nth term of an arithmetic series is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
Substituting the values, we get:
[tex]\[ a_n = 2 + (n-1) \cdot 2 = 2 + 2n - 2 = 2n \][/tex]
So, the series can be represented as:
[tex]\[ \sum_{n=1}^{50} 2n \][/tex]
where [tex]\(n\)[/tex] runs from 1 to 50.
### (c) [tex]\(1^2 + 2^2 + 4^2 + 8^2 + \cdots + 256^2\)[/tex]:
This is a geometric series where each term is a square of a power of 2. The first term [tex]\(a = 1^2 = 1\)[/tex], and the common ratio [tex]\(r = (2^2) = 4\)[/tex].
The nth term of a geometric series is given by:
[tex]\[ a_n = ar^{(n-1)} \][/tex]
Since each term is squared:
[tex]\[ a_n = (2^{(n-1)})^2 = 2^{2(n-1)} \][/tex]
So, the series can be represented as:
[tex]\[ \sum_{n=1}^{9} 2^{2(n-1)} \][/tex]
where [tex]\(n\)[/tex] runs from 1 to 9.
Therefore, the series in sigma notation are:
(a) [tex]\(\sum_{n=1}^{50} (2n - 1)\)[/tex]
(b) [tex]\(\sum_{n=1}^{50} 2n\)[/tex]
(c) [tex]\(\sum_{n=1}^{9} 2^{2(n-1)}\)[/tex]
### (a) [tex]\(1 + 3 + 5 + \cdots + 99\)[/tex]:
This is an arithmetic series with the first term [tex]\(a = 1\)[/tex] and a common difference [tex]\(d = 2\)[/tex].
The nth term of an arithmetic series is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
Substituting the values, we get:
[tex]\[ a_n = 1 + (n-1) \cdot 2 = 1 + 2n - 2 = 2n - 1 \][/tex]
So, the series can be represented as:
[tex]\[ \sum_{n=1}^{50} (2n - 1) \][/tex]
where [tex]\(n\)[/tex] runs from 1 to 50.
### (b) [tex]\(2 + 4 + 6 + \cdots + 100\)[/tex]:
This is also an arithmetic series with the first term [tex]\(a = 2\)[/tex] and a common difference [tex]\(d = 2\)[/tex].
The nth term of an arithmetic series is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
Substituting the values, we get:
[tex]\[ a_n = 2 + (n-1) \cdot 2 = 2 + 2n - 2 = 2n \][/tex]
So, the series can be represented as:
[tex]\[ \sum_{n=1}^{50} 2n \][/tex]
where [tex]\(n\)[/tex] runs from 1 to 50.
### (c) [tex]\(1^2 + 2^2 + 4^2 + 8^2 + \cdots + 256^2\)[/tex]:
This is a geometric series where each term is a square of a power of 2. The first term [tex]\(a = 1^2 = 1\)[/tex], and the common ratio [tex]\(r = (2^2) = 4\)[/tex].
The nth term of a geometric series is given by:
[tex]\[ a_n = ar^{(n-1)} \][/tex]
Since each term is squared:
[tex]\[ a_n = (2^{(n-1)})^2 = 2^{2(n-1)} \][/tex]
So, the series can be represented as:
[tex]\[ \sum_{n=1}^{9} 2^{2(n-1)} \][/tex]
where [tex]\(n\)[/tex] runs from 1 to 9.
Therefore, the series in sigma notation are:
(a) [tex]\(\sum_{n=1}^{50} (2n - 1)\)[/tex]
(b) [tex]\(\sum_{n=1}^{50} 2n\)[/tex]
(c) [tex]\(\sum_{n=1}^{9} 2^{2(n-1)}\)[/tex]
