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.