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The given sum \( S_n \) of the first \( n \) terms of the sequence \( \{u_i\}_{i=1}^{\infty} \) is:

\[
S_n = n(2n + 2024)
\]

### Part (i): Find \( u_n \) in terms of \( n \).

The sum \( S_n \) represents the sum of the first \( n \) terms of the sequence, i.e.,

\[
S_n = u_1 + u_2 + \dots + u_n.
\]

The sum of the first \( n-1 \) terms is:

\[
S_{n-1} = u_1 + u_2 + \dots + u_{n-1}.
\]

The \( n \)-th term \( u_n \) is the difference between \( S_n \) and \( S_{n-1} \), i.e.,

\[
u_n = S_n - S_{n-1}.
\]

Now, substitute the given formula for \( S_n \):

\[
S_n = n(2n + 2024),
\]

and for \( S_{n-1} \):

\[
S_{n-1} = (n-1)(2(n-1) + 2024).
\]

We calculate this difference to find \( u_n \):

\[
u_n = n(2n + 2024) - (n-1)(2(n-1) + 2024).
\]

Let's simplify this expression:

\[
u_n = n(2n + 2024) - (n-1)(2n - 2 + 2024).
\]
\[
u_n = n(2n + 2024) - (n-1)(2n + 2022).
\]
\[
u_n = n(2n + 2024) - (n-1)(2n + 2022).
\]

Simplify further:

\[
u_n = n(2n + 2024) - (n-1)(2n + 2022).
\]

You can expand and subtract these to get the final value.

Would you like more details on the simplifications or clarifications?

Here are some related questions for practice:
1. What happens to \( u_n \) for large values of \( n \)?
2. Can you find \( u_1 \) explicitly?
3. How would the sequence behave if the formula for \( S_n \) were changed to include a different constant?
4. What happens to the sequence if the coefficient of \( n \) is changed?
5. Can you deduce the general nature of the sequence (arithmetic, geometric, etc.) based on this form of \( S_n \)?

**Tip**: Breaking down complex expressions into simpler parts can make the process more manageable when calculating sequences like these.

The sum S_n of the first n terms of a sequence {u_i} is given by S_n = n(2n + 2024). (i) Find u_n in terms of n. (ii) Find u_{n+1} in terms of u_n.

Learn how to find the nth term u_n of a sequence where the sum S_n of the first n terms is given by S_n = n(2n + 2024). This problem involves algebraic manipulation and summation concepts suitable for Grades 11-12.

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