Question

solve this problem ​

Answer 1

Answer:

11

Step-by-step explanation:

Given :

To find the value of a & b if,

143

 ∑ $\frac{1}{\sqrt{k} +\sqrt{k+1} } = a - \sqrt{b}$

k = 1

Solution :

By taking conjugate ,

$\frac{1}{\sqrt{k} +\sqrt{k+1} }$

⇒ $\frac{1}{\sqrt{k} +\sqrt{k+1} }$

⇒ $\frac{1}{\sqrt{k} +\sqrt{k+1} } \times \frac{\sqrt{k} -\sqrt{k+1}}{\sqrt{k} -\sqrt{k+1} }$

⇒ $\frac{\sqrt{k} -\sqrt{k+1}}{(\sqrt{k})^2 -(\sqrt{k+1})^2}$

⇒ $\frac{\sqrt{k} -\sqrt{k+1}}{k - (k + 1)} = \frac{\sqrt{k} -\sqrt{k+1}}{-1} = \sqrt{k+1} - \sqrt{k}$

__

∴

143

 ∑ $\frac{1}{\sqrt{k} +\sqrt{k+1} }$

k = 1

=

143

 ∑ $\sqrt{k+1} - \sqrt{k}$

k = 1

⇒ ($(\sqrt{1+1} - \sqrt{1}) + (\sqrt{2+1} -\sqrt{2} ) + (\sqrt{3+1} -\sqrt{3}) + ...+ (\sqrt{142 +1} - \sqrt{142}  ) + (\sqrt{143 + 1} - \sqrt{143})$

⇒ $(\sqrt{2} - \sqrt{1}) + (\sqrt{3} -\sqrt{2} ) + (\sqrt{4} -\sqrt{3}) + ...+ (\sqrt{143} - \sqrt{142}  ) + (\sqrt{144} - \sqrt{143})$

⇒ $-\sqrt{1} + \sqrt{144} = -1 + 12 = 11$