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The sequence given is defined by the formula for each term \( b_k \) as:

\[
b_k = \frac{1}{k} - \frac{1}{k+1}
\]

for integers \( k \) from 1 to 100.

### Step-by-Step Solution:

We are asked to find the sum of the first 100 terms of this sequence, that is:

\[
S = b_1 + b_2 + \cdots + b_{100}
\]

Using the formula for \( b_k \), we have:

\[
S = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{100} - \frac{1}{101} \right)
\]

Notice that this is a **telescoping series** where most terms cancel out. Specifically, all terms in the middle cancel, and we are left with:

\[
S = \left( \frac{1}{1} - \frac{1}{101} \right)
\]

Thus, the sum is:

\[
S = 1 - \frac{1}{101} = \frac{101}{101} - \frac{1}{101} = \frac{100}{101}
\]

### Final Answer:

The sum of the 100 terms of the sequence is:

\[
S = \frac{100}{101}
\]

---

Would you like any further details or explanations?

Here are some related questions to deepen your understanding:
1. How does a telescoping series work in general?
2. Can this formula be applied for an arbitrary number of terms, like 200 terms?
3. What happens to the sum of the series as the number of terms approaches infinity?
4. Can you derive the sum formula without using the telescoping property?
5. What is the role of partial sums in calculating series like this?

**Tip:** Always look for patterns in sequences, such as cancellation in telescoping series, as it can greatly simplify the sum calculation!

In the sequence b1, b2, b3, ..., b100, the kth term is defined by bk = 1/k - 1/(k+1) for all integers k from 1 to 100. What is the sum of the 100 terms of the sequence?

This problem explores how to calculate the sum of a telescoping series, specifically for the sequence bk = 1/k - 1/(k+1) from k = 1 to 100. By simplifying the series, the sum is found to be 100/101. Learn how to approach telescoping series with detailed explanations and step-by-step solutions.

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