https://file.aiquickdraw.com/mathbot/file/1730505783036iz7t4ku6.jpg

Let’s analyze and evaluate each expression.

### Part (a)
The given expression is:
\[
\sum_{i=6}^{97} \left( \frac{1}{i} - \frac{1}{i-1} \right)
\]
This type of summation suggests **telescoping**. Here’s how it works:

1. **Expand a few terms** to observe the pattern:
   \[
   \left( \frac{1}{6} - \frac{1}{5} \right) + \left( \frac{1}{7} - \frac{1}{6} \right) + \left( \frac{1}{8} - \frac{1}{7} \right) + \dots + \left( \frac{1}{97} - \frac{1}{96} \right)
   \]

2. Notice that **many terms cancel out**:
   - For example, \( \frac{1}{6} \) in the first term cancels with \( -\frac{1}{6} \) in the second term, \( \frac{1}{7} \) cancels with \( -\frac{1}{7} \), and so forth.

3. Only the **first and last terms** remain after cancellation:
   \[
   -\frac{1}{5} + \frac{1}{97}
   \]

4. Therefore, the sum simplifies to:
   \[
   \frac{1}{97} - \frac{1}{5}
   \]

5. Calculating this result:
   \[
   \frac{1}{97} - \frac{1}{5} = \frac{5 - 97}{485} = -\frac{92}{485}
   \]

So, the answer for **(a)** is:
\[
-\frac{92}{485}
\]

---

### Part (b)
The given expression is:
\[
\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \left[ \left( \frac{4i}{n} \right)^3 + 3 \right]
\]
This limit represents a **Riemann sum**, which approximates the integral of a function over an interval as \( n \to \infty \).

1. **Rewrite the expression in terms of a Riemann sum**:
   - We can identify \( f(x) = (4x)^3 + 3 \) and the interval from 0 to 1.
   - With \( x = \frac{i}{n} \), \( \Delta x = \frac{1}{n} \), and \( \frac{4i}{n} = 4x \), the sum approximates:
     \[
     \int_0^1 \left( (4x)^3 + 3 \right) \, dx
     \]

2. **Set up the integral**:
   \[
   \int_0^1 \left( 64x^3 + 3 \right) \, dx
   \]

3. **Evaluate the integral**:
   - Separate the integral:
     \[
     \int_0^1 64x^3 \, dx + \int_0^1 3 \, dx
     \]
   - For the first integral:
     \[
     \int_0^1 64x^3 \, dx = 64 \cdot \frac{x^4}{4} \Big|_0^1 = 64 \cdot \frac{1}{4} = 16
     \]
   - For the second integral:
     \[
     \int_0^1 3 \, dx = 3x \Big|_0^1 = 3
     \]

4. **Combine the results**:
   \[
   16 + 3 = 19
   \]

So, the answer for **(b)** is:
\[
19
\]

---

Would you like further details or explanations on any step?

### Related Questions
1. How does telescoping work in summations, and when is it useful?
2. What is a Riemann sum, and how does it relate to definite integrals?
3. Why do terms cancel out in telescoping series?
4. How can you identify that a summation represents a Riemann sum?
5. What are the practical applications of Riemann sums in real-world scenarios?

### Tip
When dealing with limits of sums as \( n \to \infty \), always check if it can be converted into a Riemann sum, as this simplifies the evaluation process significantly.

Evaluate each of the following.
(a) \( \sum_{i=6}^{97} \left( \frac{1}{i} - \frac{1}{i-1} \right) \)
(b) \( \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \left[ \left( \frac{4i}{n} \right)^3 + 3 \right] \)

This page provides solutions to problems involving a telescoping series and a Riemann sum limit. Part (a) simplifies a telescoping series from i=6 to 97, while part (b) evaluates the limit of a Riemann sum that represents an integral. Suitable for students in undergraduate calculus.

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