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Powers and Roots Practice Test 4

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Question : 13 of 40

Marks: +1, -0

The value of

3+\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 equal to:

4
3
2
$3-\sqrt{8}$

Solution:

Concept:

Rationalize each term to form a telescoping series.

Explanation:

For each term, use

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

.

So,

\frac{1}{\sqrt{4}+\sqrt{5}}=\sqrt{5}-\sqrt{4}

.

Similarly,

\frac{1}{\sqrt{5}+\sqrt{6}}=\sqrt{6}-\sqrt{5}

, and so on.

The given sum becomes:

3+(\sqrt{5}-\sqrt{4})+(\sqrt{6}-\sqrt{5})+(\sqrt{7}-\sqrt{6})+(\sqrt{8}-\sqrt{7})+(\sqrt{9}-\sqrt{8})

.

All middle terms cancel out.

This leaves

3+\sqrt{9}-\sqrt{4}

.

Now,

\sqrt{9}=3

and

\sqrt{4}=2

.

So,

3+3-2=4

.

Answer:

4

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