Question

${1/√2+1}$ + ${1/√3+√2}$ + ${1/√4+√3}$ +.... ${1/√9+√8}$

Please solve this for me​

Answer 1

Answer:

$\frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{ \sqrt{4 } + \sqrt{3} } + ...... + \frac{1}{ \sqrt{9} + \sqrt{8} }$

$</p><p></p><p> = \frac{ \sqrt{2} - 1 }{ \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{1} + \frac{ \sqrt{4} - \sqrt{3} }{1} + ... + \frac{ \sqrt{9} - \sqrt{8} }{1}$

$</p><p></p><p>= \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + ... \sqrt{9} - \sqrt{8}$

$</p><p></p><p>= - 1 + \sqrt{9} \\ \\ = - 1 (+ 3) \\ \\ = 2$

$</p><p></p><p>= \frac{1}{ \sqrt{2} + \sqrt{1} } \times \frac{ \sqrt{2} - \sqrt{1} }{ \sqrt{2} - \sqrt{1} } + \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{1}{ \sqrt{4} + \sqrt{3} } \times \frac{ \sqrt{4} - \sqrt{3} }{ \sqrt{4} - 3} + \\ \frac{1}{ \sqrt{9} + \sqrt{8} } \times \frac{ \sqrt{9} - \sqrt{8} }{ \sqrt{9} - \sqrt{8} }$

$</p><p></p><p></p><p></p><p>= \frac{ \sqrt{2} - 1}{( \sqrt{2}) {}^{2} - 1} </p><p> + \frac{ \sqrt{3} </p><p> - \sqrt{2} }{( \sqrt{3}) {}^{2} </p><p> - ( \sqrt{2}) {}^{2} } </p><p>+ \frac{ \sqrt{4} - </p><p> \sqrt{3} }{( \sqrt{4} ) {}^{3} </p><p>- ( \sqrt{3} ) {}^{2} } + .... + \frac{ \sqrt{9} </p><p> - \sqrt{8} }{( \sqrt{9}) {}^{2} - ( \sqrt{8} ) {}^{2} }$

Hence, 2 is the answer.