sum of first n terms of infinite series is (3n^2+5n)/2. find un

To find the \( n \)th term (\( u_n \)) of the infinite series, we need to use the formula for the sum of the first \( n \) terms, \( S_n \), and the relationship between \( S_n \) and \( u_n \).

Given:
\[
S_n = \frac{3n^2 + 5n}{2}
\]

The \( n \)th term of the series is given by:
\[
u_n = S_n - S_{n-1}
\]

First, let's calculate \( S_{n-1} \) by substituting \( n-1 \) in place of \( n \) in the formula for \( S_n \):
\[
S_{n-1} = \frac{3(n-1)^2 + 5(n-1)}{2}
\]

Expanding \( S_{n-1} \):
\[
S_{n-1} = \frac{3((n-1)^2) + 5(n-1)}{2} = \frac{3(n^2 - 2n + 1) + 5n - 5}{2}
\]
\[
S_{n-1} = \frac{3n^2 - 6n + 3 + 5n - 5}{2} = \frac{3n^2 - n - 2}{2}
\]

Now, calculate \( u_n \) by subtracting \( S_{n-1} \) from \( S_n \):
\[
u_n = S_n - S_{n-1} = \frac{3n^2 + 5n}{2} - \frac{3n^2 - n - 2}{2}
\]

Simplify the expression:
\[
u_n = \frac{(3n^2 + 5n) - (3n^2 - n - 2)}{2}
\]
\[
u_n = \frac{3n^2 + 5n - 3n^2 + n + 2}{2} = \frac{6n + 2}{2} = 3n + 1
\]

Thus, the \( n \)th term of the series is:
\[
\boxed{3n + 1}
\]

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. How do you find the sum of an arithmetic series?
2. What is the formula for the sum of the first \( n \) terms of a geometric series?
3. How can you find the sum of an infinite series?
4. What is the difference between an arithmetic series and a geometric series?
5. How do you derive the formula for the \( n \)th term from the sum of the first \( n \) terms?

**Tip:** When dealing with series, the sum and individual terms are closely related. Understanding how to transition between these forms is key in solving series-related problems.

Learn how to calculate the nth term (un) of the infinite series using the given sum formula (3n^2 + 5n)/2. This problem covers series summation, sequences, and the derivation of the nth term from the sum of the series. Suitable for advanced high school students (Grades 11-12).

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