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The given problem involves a summation of terms of the form:

\[
\frac{1}{\sqrt{n} + \sqrt{n+1}}
\]
from \( n = 2 \) to \( n = 35 \).

### Step-by-step solution:

1. **Simplify each term**:
   \[
   \frac{1}{\sqrt{n} + \sqrt{n+1}} = \frac{1}{\sqrt{n} + \sqrt{n+1}} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2}
   \]
   The denominator simplifies to:
   \[
   (\sqrt{n+1})^2 - (\sqrt{n})^2 = (n+1) - n = 1
   \]
   So the expression becomes:
   \[
   \frac{\sqrt{n+1} - \sqrt{n}}{1} = \sqrt{n+1} - \sqrt{n}
   \]
   Thus, each term simplifies to:
   \[
   \sqrt{n+1} - \sqrt{n}
   \]

2. **Summation of terms**:
   The summation becomes:
   \[
   (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + (\sqrt{5} - \sqrt{4}) + \dots + (\sqrt{36} - \sqrt{35})
   \]
   This is a telescoping series, where most terms cancel out, leaving:
   \[
   \sqrt{36} - \sqrt{2} = 6 - \sqrt{2}
   \]

Thus, the final result is:
\[
6 - \sqrt{2}
\]

### Answer: (C) \( 6 - \sqrt{2} \)

Do you need further clarification or details on this solution?

Here are 5 related questions you might want to explore:
1. How do telescoping series work in mathematical summations?
2. What other forms of series use the method of simplifying terms like in this problem?
3. How can you generalize this type of series for any starting and ending values?
4. What is the role of rationalizing in simplifying expressions with square roots?
5. How do telescoping series apply to integral calculus problems?

**Tip:** In telescoping series, always look for cancellation patterns that simplify large sums easily.

Find the value of the following series: 1 / (√2 + √3) + 1 / (√3 + √4) + 1 / (√4 + √5) + ... + 1 / (√35 + √36)

Learn how to evaluate a telescoping series involving square roots, including simplification techniques and rationalization. The problem involves a summation from n = 2 to n = 35, resulting in the final answer of 6 - √2.

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