Sure, let's first write the sum [tex]\(3 + 5 + 7 + \ldots + 27\)[/tex] using sigma notation.
This is an arithmetic series where the first term [tex]\( a_1 \)[/tex] is [tex]\( 3 \)[/tex], the common difference [tex]\( d \)[/tex] is [tex]\( 2 \)[/tex], and the last term [tex]\( a_n \)[/tex] is [tex]\( 27 \)[/tex].
In sigma notation, the sum can be written as:
[tex]\[ \sum_{k=0}^{n-1} (3 + 2k) \][/tex]
Next, we need to determine the number of terms [tex]\( n \)[/tex] in the series.
Given the last term of the sequence (27), we can use the formula for the [tex]\( n \)[/tex]-th term of an arithmetic series:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
[tex]\[ 27 = 3 + (n-1) \cdot 2 \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ 27 = 3 + 2(n-1) \][/tex]
[tex]\[ 27 = 3 + 2n - 2 \][/tex]
[tex]\[ 27 = 2n + 1 \][/tex]
[tex]\[ 26 = 2n \][/tex]
[tex]\[ n = 13 \][/tex]
Now we can use the sum formula for an arithmetic series:
[tex]\[ S_n = \frac{n}{2} \left(a_1 + a_n\right) \][/tex]
Substitute the known values:
[tex]\[ n = 13 \][/tex]
[tex]\[ a_1 = 3 \][/tex]
[tex]\[ a_n = 27 \][/tex]
Thus, the sum [tex]\( S_n \)[/tex] is:
[tex]\[ S_n = \frac{13}{2} \left(3 + 27\right) \][/tex]
[tex]\[ S_n = \frac{13}{2} \cdot 30 \][/tex]
[tex]\[ S_n = 13 \cdot 15 \][/tex]
[tex]\[ S_n = 195 \][/tex]
Therefore, the sum of the series is:
[tex]\[ \sum_{k=0}^{12} (3 + 2k) = 195 \][/tex]
In conclusion:
- Number of terms, [tex]\( n \)[/tex] = 13.
- The sum, [tex]\( S_n \)[/tex], of the series is [tex]\( 195 \)[/tex].
