Question

prove that 1upon1+root 2+1upon root 2+root 3++1upon root 3+root 4+1upon root 4+root 5+1upon root5+root 6+1upon root 6+root 7+1upon root7+root 8+1upon root9+root 3 =2​

Answer 1

Answer:-

To Prove:

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

By rationalising the denominator a we get,

$\implies \sf\: \dfrac{1}{ \sqrt{1} + \sqrt{2} } \times \dfrac{ \sqrt{1} - \sqrt{2} }{ \sqrt{1} - \sqrt{2} } + \dfrac{1}{ \sqrt{2} + \sqrt{3} } \times \dfrac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } + \dfrac{1}{ \sqrt{3} + \sqrt{4} } \times \frac{ \sqrt{3} - \sqrt{4} }{ \sqrt{3} - \sqrt{4} } + \dfrac{1}{ \sqrt{4} + \sqrt{5} } \times \frac{ \sqrt{4} - \sqrt{5} }{ \sqrt{4} - \sqrt{5} } + \dfrac{1}{ \sqrt{5} + \sqrt{6} } \times \frac{ \sqrt{5} - \sqrt{6} }{ \sqrt{5} - \sqrt{6} } + \dfrac{1}{ \sqrt{6} + \sqrt{7} }\times \dfrac{\sqrt{6} - \sqrt{7}}{ \sqrt{6}-\sqrt{7} }+ \dfrac{1}{ \sqrt{7} + \sqrt{8} } \times \frac{ \sqrt{7} - \sqrt{8} }{ \sqrt{7} - \sqrt{8} } + \dfrac{1}{ \sqrt{8} + \sqrt{9} } \times \frac{ \sqrt{8} - \sqrt{9} }{\sqrt{8} - \sqrt{9}} = 2$

Using the identity (a + b)(a - b) = a² - b² we get,

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

Now,

By Multiplying the numerators with ( - ) we get,

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

Hence , Proved.

Answer 2

$\huge{\underline{\underline{\sf{\pink{Required\:Answer:}}}}}$

Here in this question, we are asked to prove this equation,

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

To do so, we need to

• Rationalize the denominator of each term.

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

• Here, we will use the algebraic identity $\large{\boxed{\sf{\orange{a^{2}-b^{2} = (a+b)(a-b)}}}}$

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

After that,

• Cancel the terms and simplify the equation.

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

Hence, Proved!

And we are done! :D

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