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139(A)
Найти сумму:

(1)/(1+sqrt2)+(1)/(sqrt2+sqrt3)+dots+(1)/(sqrt2006+sqrt2007)

$139(\mathrm{~A})$ $\newline$ Найти сумму: $\newline$ $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{2006}+\sqrt{2007}}$

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Multiplication with rational exponents

Full solution

Q.

139(\mathrm{~A})

\newline

Найти сумму:

\newline

\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{2006}+\sqrt{2007}}

Multiply by Conjugate: We can use the method of conjugates to simplify each term. Multiply the numerator and denominator by the conjugate of the denominator. $\newline$ For the first term: $(\frac{1}{1+\sqrt{2}}) \times (\frac{1-\sqrt{2}}{1-\sqrt{2}}) = \frac{1-\sqrt{2}}{1-2}$ .
Simplify First Term: Simplify the first term: $(1-\sqrt{2})/(-1) = \sqrt{2}-1$ .
Apply Method to All Terms: Apply the same method to all terms in the series. Each term will look like $(\sqrt{n+1}-\sqrt{n})$ .
Telescoping Effect: Notice that when we add the terms, there will be a telescoping effect. Most terms will cancel out, leaving only the first and last parts of the square roots from the first and last terms.
Add Simplified Terms: Add the simplified terms: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + \ldots + (\sqrt{2007}-\sqrt{2006})$ .
Final Cancellation: After cancellation, we're left with: $-1 + \sqrt{2007}.$
Final Answer: The final answer is $\sqrt{2007} - 1$ .

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