Question

the value of
$\frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3 } } + \frac{1}{ \sqrt{3} + \sqrt{4} } + \frac{1}{\sqrt{4 } + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{6} } + \frac{1}{ \sqrt{6} + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{8} } + \frac{1}{ \sqrt{8} + \sqrt{9} }$
​

Answer 1

Topic

Rationalization

• When there is a sum or difference of square roots, we can simply remove the surd by squaring each term. The squaring of each term is done by the rationalization factor.
• Polynomial identity $(a+b)(a-b)=a^{2}-b^{2}$ squares each term.

Telescoping series

• When there are repeating terms with different signs, it would remove both values. This results in the cancellation, except for the start and the end.

Solution

To avoid confusion, let's write the denominator in descending order.

$\dfrac{1}{\sqrt{2} +1} +\dfrac{1}{\sqrt{3} +\sqrt{2} } +...+\dfrac{1}{\sqrt{9} +\sqrt{8} }$

Before we calculate let's take an example.

Example 1:-

$\implies (\sqrt{2} +1)(\sqrt{2} -1)=(\sqrt{2} )^{2}-1^{2}=\boxed{1}$

The rationalization factor of the first fraction is $\sqrt{2} -1$.

After simplification, we get $\dfrac{\sqrt{2} -1}{(\sqrt{2} +1)(\sqrt{2} -1)}=\dfrac{\sqrt{2} -1}{1} =\boxed{\sqrt{2} -1}$.

Example 2:-

$\implies (\sqrt{3} +\sqrt{2} )(\sqrt{3} -\sqrt{2} )=(\sqrt{3} )^{2}-(\sqrt{2} )^{2}=\boxed{1}$

The next rationalization factor is $\sqrt{3} -\sqrt{2}$.

After simplification, we get $\dfrac{\sqrt{3} -\sqrt{2} }{(\sqrt{3} +\sqrt{2} )(\sqrt{3} -2)}=\dfrac{\sqrt{3} -\sqrt{2} }{1} =\boxed{\sqrt{3} -\sqrt{2} }$.

"What do we see here?"

We can see that the denominator equals always 1. It is because the square of the surds differ by 1. We also see that the numerator is a difference of two surds.

However, if we add all the fraction the result comes out as $\sqrt{9} -\sqrt{1}$, because the terms between the first and the last term cancel out.

$\implies (\cancel{\sqrt{2} }-\sqrt{1} )+(\cancel{\sqrt{3} }-\cancel{\sqrt{2} })+...+(\sqrt{9} -\cancel{\sqrt{8} })=\boxed{\sqrt{9} -\sqrt{1} }$

After we simplify this we get $\boxed{2}$ as the answer.

This is the reason this is called a telescoping series.

Answer 2

Step-by-step explanation:

QᴜᴇsᴛɪᴏN࿐Parapodia is the fleshy protrusion found on the marine gastropods. It is used for locomotion and for respiration. Sea snails and sea slugs have parapodia that is used for swimming.