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The given expression involves summing the function \( \frac{4i + 9}{n^2} \) from \( i = 1 \) to \( n \). Let’s break it down step by step to rewrite the summation and then calculate the sum for the values of \( n = 10, 100, 1000, \) and \( 10000 \).

### Step 1: Rewrite the summation

The summation formula given is:

\[
S(n) = \sum_{i=1}^{n} \frac{4i + 9}{n^2}
\]

This can be split into two separate sums:

\[
S(n) = \sum_{i=1}^{n} \frac{4i}{n^2} + \sum_{i=1}^{n} \frac{9}{n^2}
\]

Now, let’s handle each sum individually.

#### First sum: \( \sum_{i=1}^{n} \frac{4i}{n^2} \)

\[
\sum_{i=1}^{n} \frac{4i}{n^2} = \frac{4}{n^2} \sum_{i=1}^{n} i
\]

We know the formula for the sum of the first \( n \) integers is:

\[
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}
\]

So:

\[
\sum_{i=1}^{n} \frac{4i}{n^2} = \frac{4}{n^2} \cdot \frac{n(n+1)}{2} = \frac{2(n+1)}{n}
\]

#### Second sum: \( \sum_{i=1}^{n} \frac{9}{n^2} \)

Since \( \frac{9}{n^2} \) is constant with respect to \( i \), it just multiplies by \( n \):

\[
\sum_{i=1}^{n} \frac{9}{n^2} = \frac{9n}{n^2} = \frac{9}{n}
\]

### Step 2: Combine the two sums

Now combine the results of both sums:

\[
S(n) = \frac{2(n+1)}{n} + \frac{9}{n}
\]

Simplifying further:

\[
S(n) = \frac{2n + 2 + 9}{n} = \frac{2n + 11}{n}
\]

Thus, the final expression for \( S(n) \) is:

\[
S(n) = 2 + \frac{11}{n}
\]

### Step 3: Compute \( S(n) \) for specific values of \( n \)

Now, let’s plug in the given values of \( n \):

1. **For \( n = 10 \):**

\[
S(10) = 2 + \frac{11}{10} = 2 + 1.1 = 3.1
\]

2. **For \( n = 100 \):**

\[
S(100) = 2 + \frac{11}{100} = 2 + 0.11 = 2.11
\]

3. **For \( n = 1000 \):**

\[
S(1000) = 2 + \frac{11}{1000} = 2 + 0.011 = 2.011
\]

4. **For \( n = 10000 \):**

\[
S(10000) = 2 + \frac{11}{10000} = 2 + 0.0011 = 2.0011
\]

### Final Answers:

- \( S(10) = 3.1 \)
- \( S(100) = 2.11 \)
- \( S(1000) = 2.011 \)
- \( S(10000) = 2.0011 \)

Would you like any more details or explanations?

### Relative Questions:
1. How can you break down a more complex summation involving multiple terms?
2. What are the common summation formulas and how are they applied?
3. Why do terms like \( \sum i \) follow specific patterns?
4. How does the behavior of the summation change as \( n \) grows larger?
5. How do small terms like \( \frac{11}{n} \) influence the sum for large values of \( n \)?

### Tip:
When working with summations, breaking them down into smaller, more manageable parts often simplifies the calculation.

Use the summation formulas to rewrite the expression ∑ (4i + 9) / n² without the summation notation. Find the sums for n = 10, 100, 1000, and 10,000.

Learn how to rewrite the summation ∑ (4i + 9) / n² and compute the sums for n = 10, 100, 1000, and 10,000 using basic summation formulas. This problem covers algebraic summation concepts suitable for students in Grades 10-12.

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