Question

prove that
$\binom{1}{1 + \sqrt{2} } + \binom{1}{ \sqrt{2} + \sqrt{3} } + \binom{1}{ \sqrt{3} + \sqrt{4} }...+ \binom{1}{ \sqrt{8} + 3} = 2$
​

Answer 1

Given:-

• $\dfrac{1}{1 + \sqrt{2} } + \dfrac{1}{ \sqrt{2} + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + \sqrt{4} } + \dots + \dfrac{1}{ \sqrt{8} + 3} = 2$

We have to proof this.

Proof:-

Considering L.H.S.:

First look at each term individually. We observe that we can rationalise the denominator of each of the term by multiplying the denominator and numerator with the conjugate pair of the denominator.

1st term:

$\dfrac{1}{1 + \sqrt{2} } = \dfrac{1(1 - \sqrt{2}) }{(1 + \sqrt{2})(1 - \sqrt{2} ) }$

Expanding denominator with (a + b)(a - b) = a² - b² with a = 1 and b = √2,

$\longrightarrow \dfrac{1}{1 + \sqrt{2} } = \dfrac{1 - \sqrt{2} }{ {(1)}^{2} - {( \sqrt{2}) }^{2}}$

$\longrightarrow \dfrac{1}{1 + \sqrt{2} } = \dfrac{1 - \sqrt{2} }{ - 1}$

$\longrightarrow \dfrac{1}{1 + \sqrt{2} } = - (1 - \sqrt{2})$

Repeating the same steps for next terms, we get,

$\longrightarrow \dfrac{1}{ \sqrt{2} + \sqrt{3} } = - ( \sqrt{2} - \sqrt{3})$

Next term:

$\longrightarrow \dfrac{1}{ \sqrt{3} + \sqrt{4} } = - ( \sqrt{3} - \sqrt{4})$

And repeating the same steps for the last term, we get:

$\longrightarrow \dfrac{1}{ \sqrt{8} + 3 } = - ( \sqrt{8} - 3)$

So, on adding them, we get,

$= - (1 - \sqrt{2} ) - ( \sqrt{2} - \sqrt{3} ) - ( \sqrt{3} - \sqrt{4} ) \: - \dots \dots \: - ( \sqrt{8} - 3)$

$= - 1 + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{4} \: - \dots \dots \: - \sqrt{8} + 3$

$= - 1 +   \cancel{ \sqrt{2} } - \cancel{ \sqrt{2} } + \cancel{ \sqrt{3} } - \cancel{ \sqrt{3} } + \cancel{ \sqrt{4} } \:  \:  -   \cancel{ \dots} \dots \cancel{ \dots} \:\:-  \cancel{ \sqrt{8} } +  3$ (refer to attachment.)

$= - 1 + 3$

$= 2$

Considering R.H.S.:

$2$

As L.H.S. = R.H.S.,

hence proved!