Question

Question 1
Prove that
$\sf \dashrightarrow \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}}$

Question 2
‎‎Prove that

$\sf \dashrightarrow \ \dfrac{3 + 2\sqrt{2}}{2 - \sqrt{2}} - \dfrac{3 - 2\sqrt{2}}{2 + \sqrt{2}} = a + b\sqrt{2}$

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Answer 1

Answer:

1 ) 2

2 ) a = 0 ; b = 7

Step-by-step explanation:

Appropriate Question :

Question 1

$\mathrm{ \dashrightarrow \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}}}$

Question 2

$\mathrm{\dashrightarrow \ \dfrac{3 + 2\sqrt{2}}{2 - \sqrt{2}} - \dfrac{3 - 2\sqrt{2}}{2 + \sqrt{2}} = a + b\sqrt{2}}$

To find :

1 ) Value of the given expression

2 ) Value of a and b

Solution :

❖ Question 1

$\mathrm{ \dashrightarrow \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}}}$

Rationalizing each term

We know that,$\boxed{\mathrm{if\ \dfrac{1}{\sqrt a - \sqrt b} \ then\ for\ rationalizing\ we\ do\ \dfrac{\sqrt a + \sqrt b}{(\sqrt a- \sqrt b)(\sqrt a + \sqrt b)} = \dfrac{\sqrt a + \sqrt b}{a - b} }}$

$\mathrm{ \implies \dfrac{\sqrt{2} - \sqrt{1}}{2-1} + \dfrac{\sqrt{3} - \sqrt{2}}{3-2} + \dfrac{\sqrt{4} - \sqrt{3}}{4-3} + {\dots\dots\dots} \ + \dfrac{\sqrt{9} - \sqrt{8}}{9-8}}$

$\mathrm{ \implies \dfrac{\sqrt{2} - \sqrt{1}}{1} + \dfrac{\sqrt{3} - \sqrt{2}}{1} + \dfrac{\sqrt{4} - \sqrt{3}}{1} + {\dots\dots\dots} \ + \dfrac{\sqrt{9} - \sqrt{8}}{1}}$

$\implies \mathrm{\cancel{\sqrt{2}} - 1 + \cancel{\sqrt{3}} - \cancel{\sqrt{2}} + \cancel{\sqrt{4}} - \cancel{\sqrt{3}} + \dots \dots\dots + \sqrt{9} - \cancel{\sqrt{8}}}$

$\implies \mathrm{\sqrt{9} - 1}$

$\implies 3 - 1$

$\therefore \underline{\boxed{\mathbf{ \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}} = \huge{\text{2}}}}}$

2 is the answer for this question.

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❖ Question 2

$\mathrm{\dashrightarrow \ \dfrac{3 + 2\sqrt{2}}{2 - \sqrt{2}} - \dfrac{3 - 2\sqrt{2}}{2 + \sqrt{2}} = a + b\sqrt{2}}$

Equalizing the denominators, we get,

$\mathrm{\implies \ \dfrac{(3 + 2\sqrt{2})(2 + \sqrt 2)}{4 -2} - \dfrac{(3 - 2\sqrt{2})(2 - \sqrt 2)}{4 - 2} = a + b\sqrt{2}}$

$\mathrm{\implies \ \dfrac{(3 + 2\sqrt{2})(2 + \sqrt 2) - (3 - 2\sqrt{2})(2 - \sqrt 2)}{4 -2} = a + b\sqrt{2}}$

$\mathrm{\implies \ \dfrac{(6 + 3\sqrt 2+ 4\sqrt{2} + 4) - (6 - 3\sqrt{2}+4 -4\sqrt 2)}{2} = a + b\sqrt{2}}$

$\mathrm{\implies \ \dfrac{6 + 3\sqrt 2+ 4\sqrt{2} + 4 - 6 + 3\sqrt{2}-4 +4\sqrt 2}{2} = a + b\sqrt{2}}$

$\implies \mathrm{\dfrac{14\sqrt 2}{2} = a + b\sqrt2}$

$\implies \mathrm{7\sqrt2 = a + b\sqrt2}$

So comparing on both sides, we get,

$\boxed{\mathrm{a = 0\ ; \ b = 7}}$

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Hope it helps!