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

$\huge{\underline{\boxed{\red{\mathcal{ANSWER}}}}}$

To Prove:

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

Proof:

Now we need to rationalise each and every fraction.

$1) \dfrac{1}{1+\sqrt{2}}*\dfrac{1-\sqrt2}{1-\sqrt2} = \dfrac{1-\sqrt2}{(1+\sqrt{2})(1-\sqrt{2})} = \dfrac{1-\sqrt2}{1-2} = \dfrac{1-\sqrt2}{-1} = \bf{\sqrt2 - 1}$

$2) \dfrac{1}{\sqrt2+\sqrt3}*\dfrac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3} = \dfrac{\sqrt2-\sqrt3}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)} = \dfrac{\sqrt2-\sqrt3}{2-3} = \dfrac{\sqrt2-\sqrt3}{-1} = \bf{\sqrt3-\sqrt2}$

$3) \dfrac{1}{\sqrt3+\sqrt4}*\dfrac{\sqrt3-\sqrt4}{\sqrt3-\sqrt4} = \dfrac{\sqrt3-\sqrt4}{(\sqrt3+\sqrt4)(\sqrt3-\sqrt4)} = \dfrac{\sqrt3-\sqrt4}{3-4} = \dfrac{\sqrt3-\sqrt4}{-1} = \bf{\sqrt4-\sqrt3}$

$4) \dfrac{1}{\sqrt4+\sqrt5}*\dfrac{\sqrt4-\sqrt5}{\sqrt4-\sqrt5} = \dfrac{\sqrt4-\sqrt5}{(\sqrt4+\sqrt5)(\sqrt4-\sqrt5)} = \dfrac{\sqrt4-\sqrt5}{4-5} = \dfrac{\sqrt4-\sqrt5}{-1} = \bf{\sqrt5-\sqrt4}$

$5) \dfrac{1}{\sqrt5+\sqrt6}*\dfrac{\sqrt5-\sqrt6}{\sqrt5-\sqrt6} = \dfrac{\sqrt5-\sqrt6}{(\sqrt5+\sqrt6)(\sqrt5-\sqrt6)} = \dfrac{\sqrt5-\sqrt6}{5-6} = \dfrac{\sqrt5-\sqrt6}{-1} = \bf{\sqrt6-\sqrt5}$

$6) \dfrac{1}{\sqrt6+\sqrt7}*\dfrac{\sqrt6-\sqrt7}{\sqrt6-\sqrt7} = \dfrac{\sqrt6-\sqrt7}{(\sqrt6+\sqrt7)(\sqrt6-\sqrt7)} = \dfrac{\sqrt6-\sqrt7}{6-7} = \dfrac{\sqrt6-\sqrt7}{-1} = \bf{\sqrt7-\sqrt6}$

$7) \dfrac{1}{\sqrt7+\sqrt8}*\dfrac{\sqrt7-\sqrt8}{\sqrt7-\sqrt8} = \dfrac{\sqrt7-\sqrt8}{(\sqrt7+\sqrt8)(\sqrt7-\sqrt8)} = \dfrac{\sqrt7-\sqrt8}{7-8} = \dfrac{\sqrt7-\sqrt8}{-1} = \bf{\sqrt8-\sqrt7}$

$8) \dfrac{1}{\sqrt8+\sqrt9}*\dfrac{\sqrt8-\sqrt9}{\sqrt8-\sqrt9} = \dfrac{\sqrt8-\sqrt9}{(\sqrt8+\sqrt9)(\sqrt8-\sqrt9)} = \dfrac{\sqrt8-\sqrt9}{8-9} = \dfrac{\sqrt8-\sqrt9}{-1} = \bf{\sqrt9-\sqrt8}$

Now, we put all the rationalised numbers in the LHS,

So,

$:\implies (\sqrt2-1) + (\sqrt3-\sqrt2) + (\sqrt4-\sqrt3) + (\sqrt5-\sqrt4) + (\sqrt6-\sqrt5) + (\sqrt7-\sqrt6) + (\sqrt8-\sqrt7) + (\sqrt9-\sqrt8)\\\\:\implies \sqrt2\!\!\!\!/\,\,-1+\sqrt3\!\!\!\!/\,\,-\sqrt2\!\!\!\!/\,\,+\sqrt4\!\!\!\!/\,\,-\sqrt3\!\!\!\!/\,\,+\sqrt5\!\!\!\!/\,\,-\sqrt4\!\!\!\!/\,\,+\sqrt6\!\!\!\!/\,\,-\sqrt5\!\!\!\!/\,\,+\sqrt7\!\!\!\!/\,\,-\sqrt6\!\!\!\!/\,\,+\sqrt8\!\!\!\!/\,\,-\sqrt7\!\!\!\!/\,\,+\sqrt9-\sqrt8\!\!\!\!/$

$:\implies 3 - 1 \:\:\:\:\:\:\:\:\:\:\:\:(\sqrt9=3)\\\bf{:\implies 2 = RHS}$

HENCE PROVED