What is the value of the following sum?
1/(√2 + √1) + 1/(√3 + √2) + 1/(√4 + √3) + ... + 1/(√100 + √99)

To solve this sum, we rationalize each term. For a general term 1/(√(n+1) + √n), we multiply the numerator and denominator by the conjugate (√(n+1) - √n):

1/(√(n+1) + √n) × (√(n+1) - √n)/(√(n+1) - √n) = (√(n+1) - √n) / ((n+1) - n) = √(n+1) - √n

Applying this to the given series, we get a telescoping series:

• Term 1: √2 - √1
• Term 2: √3 - √2
• Term 3: √4 - √3
• ...
• Term 99: √100 - √99

When we sum these terms, all intermediate values cancel out:

Sum = (√2 - √1) + (√3 - √2) + (√4 - √3) + ... + (√100 - √99)
Sum = -√1 + √100
Sum = -1 + 10 = 9.

Thus, the value of the sum is 9.