To find the value of the series [tex]\(\sum_{n=1}^{25}(3n-2)\)[/tex], we need to evaluate the sum of each term of the form [tex]\(3n - 2\)[/tex] from [tex]\(n = 1\)[/tex] to [tex]\(n = 25\)[/tex].
First, let's write down the expression for each term when [tex]\(n\)[/tex] varies from 1 to 25:
[tex]\[
\begin{align*}
\text{When } n = 1, & \quad 3 \cdot 1 - 2 = 1 \\
\text{When } n = 2, & \quad 3 \cdot 2 - 2 = 4 \\
\text{When } n = 3, & \quad 3 \cdot 3 - 2 = 7 \\
\text{When } n = 4, & \quad 3 \cdot 4 - 2 = 10 \\
&\; \vdots \\
\text{When } n = 25, & \quad 3 \cdot 25 - 2 = 73 \\
\end{align*}
\][/tex]
Next, let's find the sum of all these terms:
[tex]\[
\sum_{n=1}^{25}(3n-2)
\][/tex]
We can separate this into two separate sums:
[tex]\[
\sum_{n=1}^{25} 3n - \sum_{n=1}^{25} 2
\][/tex]
First, we calculate [tex]\(\sum_{n=1}^{25} 3n\)[/tex]:
[tex]\[
3 \sum_{n=1}^{25} n = 3 \cdot \left(\frac{25 \cdot (25 + 1)}{2}\right) = 3 \cdot \left(\frac{25 \cdot 26}{2}\right) = 3 \cdot 325 = 975
\][/tex]
Next, we calculate [tex]\(\sum_{n=1}^{25} 2\)[/tex]:
[tex]\[
25 \cdot 2 = 50
\][/tex]
Now, we combine these results to find the total sum:
[tex]\[
975 - 50 = 925
\][/tex]
Thus, the value of [tex]\(\sum_{n=1}^{25}(3n-2)\)[/tex] is:
[tex]\[
\boxed{925}
\][/tex]
