Math Problem Statement

The value of   $\small \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}$ is:

Solution

We are asked to find the value of the following sum:

S=11+2+12+3+13+4+14+5+15+6+16+7+17+8+18+9.S = \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}.

Step 1: Rationalizing the denominators

Each term in the sum has the form 1n+n+1\frac{1}{\sqrt{n} + \sqrt{n+1}}, where nn is an integer. To simplify this, we rationalize the denominator by multiplying both the numerator and the denominator by n+1n\sqrt{n+1} - \sqrt{n}.

For example, for the first term 11+2\frac{1}{1 + \sqrt{2}}, we multiply the numerator and denominator by 21\sqrt{2} - 1:

11+2=11+2×2121=21(2)212=2121=21.\frac{1}{1 + \sqrt{2}} = \frac{1}{1 + \sqrt{2}} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1.

Similarly, for the second term 12+3\frac{1}{\sqrt{2} + \sqrt{3}}, we multiply by 32\sqrt{3} - \sqrt{2}:

12+3=32(3)2(2)2=3232=32.\frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}.

Step 2: Applying this to all terms

We can apply this rationalization to all terms in the sum. The general term will look like this:

1n+n+1=n+1n.\frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}.

So, each term in the sum becomes:

11+2=21,12+3=32,13+4=43,\frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1, \quad \frac{1}{\sqrt{2} + \sqrt{3}} = \sqrt{3} - \sqrt{2}, \quad \frac{1}{\sqrt{3} + \sqrt{4}} = \sqrt{4} - \sqrt{3}, \quad \dots

Thus, the sum becomes:

S=(21)+(32)+(43)+(54)+(65)+(76)+(87)+(98).S = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + (\sqrt{5} - \sqrt{4}) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (\sqrt{8} - \sqrt{7}) + (\sqrt{9} - \sqrt{8}).

Step 3: Simplifying the sum

Notice that this is a telescoping series. Most terms cancel out with adjacent terms:

S=1+9=1+3=2.S = -1 + \sqrt{9} = -1 + 3 = 2.

Thus, the value of the sum is:

2.\boxed{2}.


Would you like further explanation or any other details? Here are some related questions:

  1. How does the telescoping property help simplify sums like this?
  2. What happens if we have a different number of terms in the sum?
  3. Can this method be applied to sums with different types of radicals?
  4. How do we rationalize denominators in general when dealing with square roots?
  5. What other types of sums can be simplified using similar techniques?

Tip: Always look for cancellation patterns (like telescoping) when summing terms that have similar structures.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rationalization of Denominators
Telescoping Series
Radicals

Formulas

Rationalizing the denominator: \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}
Sum of a telescoping series: \sum_{n} (a_{n} - a_{n+1}) = a_1 - a_n

Theorems

Rationalization of square roots
Telescoping Series

Suitable Grade Level

Grades 9-12