Question

prove that : 1/1+root 2 + 1/root2+root3 +1/root3 +root4 + ...................+ 1/root8+root9 = 2

Answer 1

1/(1+root2)+1/(root2+root3)+1/(root 3+root4)+......1/(root8+root9)
the above question proceed by first step you rationalize all fractional irrational number.
i.e
( roo2-1)+(root3-root2)+(root4-root3)+.....
(root9-root8)=3-1=2

Answer 2

The value of alpha square minus beta square is 2

Step-by-step explanation:

LHS:

$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}$

Rationalizing the denominator, we get

$\Rightarrow\left(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}+\sqrt{4}} \times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\right)+\cdots \ldots+\left(\frac{1}{\sqrt{8}+\sqrt{9}} \times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}\right)$

We know that,

$\left(a^{2}-b^{2}\right)=(a+b)(a-b)$

Now, on substituting the formula, we get,

$=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\cdots \ldots \cdot \frac{(\sqrt{8}-\sqrt{9})}{8-9}$

$\Rightarrow \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{8}+\sqrt{9}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})$

In the above step -1 in the denominator will be carried to the numerator.

On cancelling the negative and positive terms, we get,

$\Rightarrow(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})=\sqrt{9}-1$

$=3-1$

$\therefore \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}=2$

Hence proved.