Question

prove that 1/1+√2 + 1/√2+√3 + 1/√3+√4 + 1/√4+√5 + 1/√5+√6 + 1/√6+√7 + 1/√7+√8 + 1/√8+√9=2​

Answer 1

Before solving:-

First, we observe the same rule in each term. Such numbers are called sequences, and 'defining the sequences' means finding a common rule then writing in terms of a letter.

The sum of the sequence is called series, which is often denoted by $\displaystyle \sum^{n}_{k=1}a_{k}$. $k=1$ is the number for the first term, and $n$ is the number for the last term.

Solution:-

Let's define the sequence as $a_{n}=\dfrac{1}{\sqrt{n} +\sqrt{n+1} }$.

$\rightarrow a_{n}=\dfrac{1}{\sqrt{n} +\sqrt{n+1} } \times \dfrac{\sqrt{n} -\sqrt{n+1} }{\sqrt{n} -\sqrt{n+1} }$

$\rightarrow a_{n}=\dfrac{\sqrt{n} -\sqrt{n+1} }{n-(n+1)}$

$\rightarrow a_{n}=\dfrac{\sqrt{n} -\sqrt{n+1} }{-1}$

$\rightarrow a_{n}=\sqrt{n+1} -\sqrt{n}$

Then $\displaystyle \sum^{8}_{k=1}a_{k}$ is a telescoping series that cancel out, leaving the first and the last term.

$\displaystyle \sum^{8}_{k=1}a_{k}$

$=\sqrt{8+1} -\cancel{\sqrt{8} }+\cancel{\sqrt{7+1} }- ...-\cancel{\sqrt{2} }+\cancel{\sqrt{1+1} }-\sqrt{1}$

$=\sqrt{9} -\sqrt{1}$

$=3-1=\boxed{2}$

Hence proven.

Answer 2

${ \sf{ \dfrac{1}{1 + \sqrt{2} } + \dfrac{1}{ \sqrt{2 } + \sqrt{3} } + \dfrac{1}{\sqrt{3} + \sqrt{4} } + \dfrac{1}{ \sqrt{4} + \sqrt{5} } + \dfrac{1}{ \sqrt{5} + \sqrt{6} } + \dfrac{1}{ \sqrt{6} + \sqrt{7} } + \dfrac{1}{ \sqrt{7} + \sqrt{8} } + \dfrac{1}{ \sqrt{8} + \sqrt{9} } = 2}}$

Now Rationalize the L.H.S term

${ \sf{ \dfrac{1( \sqrt{2} - 1) }{(1 + \sqrt{2} )( \sqrt{2} - 1)} + \dfrac{1( \sqrt{3} - \sqrt{2} )}{ (\sqrt{2 } + \sqrt{3} )( \sqrt{3} - \sqrt{2} )} + \dfrac{1( \sqrt{4} - \sqrt{3} )}{(\sqrt{3} + \sqrt{4})( \sqrt{4} - \sqrt{3}) } + \dfrac{1( \sqrt{5} - \sqrt{4}) }{ (\sqrt{4} + \sqrt{5} )( \sqrt{5} - \sqrt{4} ) } + \dfrac{1( \sqrt{6} - \sqrt{5} )}{ (\sqrt{5} + \sqrt{6} )( \sqrt{6} - \sqrt{5}) } + \dfrac{1( \sqrt{7} - \sqrt{6} )}{ (\sqrt{6} + \sqrt{7} )( \sqrt{7} - \sqrt{6}) } + \dfrac{1( \sqrt{8} - \sqrt{7} )}{( \sqrt{7} + \sqrt{8} )( \sqrt{8} - \sqrt{7} )} + \dfrac{1( \sqrt{9} - \sqrt{8} )}{ (\sqrt{8} + \sqrt{9} )( \sqrt{9} - \sqrt{8}) } }}$

Using formula = ( a - b )( a + b ) = a² - b²

${ \sf{ { \cancel{ \sqrt{ {2}}} } - 1 + { \cancel{ \sqrt{3} }}- { \cancel{\sqrt{2} }}+ { \cancel{\sqrt{4} }}- { \cancel{\sqrt{3} }}+ { \cancel{\sqrt{5}}} - { \cancel{ \sqrt{4} }}+ { \cancel{\sqrt{6} }} - { \cancel{\sqrt{5}}} +{ \cancel{ \sqrt{7} }} - { \cancel{ \sqrt{6}}} + { \cancel{\sqrt{8}}} - { \cancel{ \sqrt{7}}} + \sqrt{9} - { \cancel{\sqrt{8} }}}}$

${ \sf{ \sqrt{9} - 1}} = 3 - 1$

${ \sf{ \therefore \: L.H.S\: \: = \: {2}}}$

L.H.S = R.H.S