Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

༒Other best user


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

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

plz No Spam. ​

Answer 1

Answer:

• Conjugate of 1 + √2 is 1 - √2
• Conjugate of √2 + √3 = √2 - √3
• Conjugate of √3 + √4 = √3 - √4
• Conjugate of √8 + √9 = √8 - √9

‎

‎

$\sf \dashrightarrow \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}} =2\\ \\ \sf \dashrightarrow \dfrac{1}{1 + \sqrt{2}} \times \dfrac{1 -\sqrt{2}}{1 -\sqrt{2}} \ + \ \dfrac{1}{\sqrt{2} + \sqrt{3}} \times \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\ \\ \\ {\ \ \ \ \ \ \ \ \ + \ \dfrac{1}{\sqrt{3} + \sqrt{4}} \times \dfrac{\sqrt{3} - \sqrt{4}}{\sqrt{3} - \sqrt{4}} \ + \ {\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}} \times \dfrac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}}= 2}$

[Broken into two lines for readability]

‎‎

Using (a + b)(a - b) = a² - b² we get;

‎‎

‎‎

$\sf \dashrightarrow \ \dfrac{1 -\sqrt{2}}{(1)^2 - (\sqrt{2})^2} + \dfrac{\sqrt{2} - \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} + \dfrac{\sqrt{3} - \sqrt{4}}{(\sqrt{3})^2 - (\sqrt{4})^2} + \ ... \ + \ \dfrac{\sqrt{8} - \sqrt{9}}{(\sqrt{8})^2 - (\sqrt{9})^2} = 2$

‎‎

$\sf \dashrightarrow \dfrac{1 -\sqrt{2}}{1 - 2} \ + \ \dfrac{\sqrt{2} - \sqrt{3}}{2 - 3} \ + \ \dfrac{\sqrt{3} - \sqrt{4}}{3 - 4} + \ {\dots} \ + \ \dfrac{\sqrt{8} - \sqrt{9}}{8 - 9} = 2\\ \\ \sf \dashrightarrow \dfrac{1 -\sqrt{2}}{-1} \ + \ \dfrac{\sqrt{2} - \sqrt{3}}{-1} \ + \ \dfrac{\sqrt{3} - \sqrt{4}}{-1} + \ {\dots} \ + \ \dfrac{\sqrt{8} - \sqrt{9}}{-1} = 2\\ \\ \sf \dashrightarrow \ \dfrac{1 -\sqrt{2} + \sqrt{2} - \sqrt{3} + \sqrt{3} - \sqrt{4} + \dots + \sqrt{8} - \sqrt{9}}{-1} = 2$

‎‎

Everything except for 1 and -√9 gets cancelled.

‎

$\sf \dashrightarrow \ \dfrac{1 - \sqrt{9}}{-1} = 2\\ \\ \sf \dashrightarrow \ \dfrac{1 - 3}{-1} = 2\\ \\\sf \dashrightarrow \ \dfrac{-2}{-1} = 2\\ \\ \sf \dashrightarrow \red{ \bold{2 = 2}}$

‎

• LHS = RHS

Hence proved.

‎

______________________________

‎______________________________

Question 2:-

‎‎

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

‎‎

On rationalizing the denominator we get;

‎‎

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

‎‎

Using (a - b)(a + b) = a² - b² in the denominators of both fractions we get;

‎

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

‎

$\sf \dashrightarrow \ \dfrac{6 + 3\sqrt{2} + 4\sqrt{2} + 2(\sqrt{2})^2}{4 - 2} - \dfrac{6 - 3\sqrt{2} - 4\sqrt{2} + 2(\sqrt{2})^2}{4 - 2} = a + b\sqrt{2}$

‎

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

‎

$\sf \dashrightarrow \ \dfrac{3\sqrt{2} + 4\sqrt{2} + 3\sqrt{2} + 4\sqrt{2}}{2} = a + b\sqrt{2}\\ \\ \sf \dashrightarrow \ \dfrac{14\sqrt{2}}{2} = a + b\sqrt{2}\\ \\ \sf \dashrightarrow \ 7\sqrt{2} = a + b\sqrt{2}$

∴ A= 0 and B = 7.

Answer 2

Answer:

Step-by-step explanation:

(i):      In general:

$\sf{\implies \frac{1}{\sqrt n + \sqrt{n+1}} +\frac{1}{\sqrt{n+1} + \sqrt{n+2}} + ...}\\\\\sf{\implies\big( \frac{1}{\sqrt n + \sqrt{n+1}} \times \frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n} -\sqrt{n+1}}\big)+\big( \frac{1}{\sqrt{n+1} + \sqrt{n+2}}\times\frac{\sqrt{n+1}-\sqrt{n+2}}{\sqrt{n+1}-\sqrt{n+2}}\big) + ...}\\\\\sf{\implies \frac{\sqrt{n}-\sqrt{n+1}}{n-(n-1)}+\frac{\sqrt{n+1}-\sqrt{n+2}}{(n+1)-(n+2)} +... }}\\\\\sf{\implies(\sqrt{n+1}-\sqrt{n})+(\sqrt{n+2}-\sqrt{n+1})... }$

         Similarly,  

$\sf{\implies\frac{1}{1+\sqrt2} + \frac{1}{\sqrt2 + \sqrt3} +...}\\\\\implies\sf{(\sqrt2-\sqrt1)+(\sqrt{3}-\sqrt2)+...+(\sqrt{n+1}-\sqrt{n}}\\\\\implies \sf{\sqrt{n+1} -1}$

     where n is the number of such fractions.

(ii):

  $\sf{\frac{1+2+2\sqrt2}{\sqrt2(\sqrt2 - 1)} - \frac{1+2-2\sqrt2}{\sqrt2(\sqrt2 + 1)} }\\\\\implies\sf{\frac{1}{\sqrt{2}}\bigg[\frac{(\sqrt2 +1)^2}{\sqrt2 -1} - \frac{(\sqrt{2}-1)^2}{\sqrt2 + 1}}\bigg]\\\\\implies \sf{\frac{1}{\sqrt2} \bigg[(\sqrt2 +1)^3 - (\sqrt2 -1)^3}\bigg]\\\\\implies \sf{ \frac{1}{\sqrt{2}} (14)}\\\\\implies\sf{7\sqrt2}$

On comparing, we get a = 0 , b = 7