Question

Prove that 1/1+√2+1/√2+√3+1/√3+√4+1/√4+√5+1/√5+√6+1/√6+√7+1/√7+√8+1/√8+√9 =2

Answer 1

Answer:

Hey mate! refer to the explanation given in the pic:

Hope it helps!

Answer 2

Rationalization

Definition:

Rationalization is a procedure used in elementary mathematics to remove the irrational number from the denominator.

Explanation:

On rationalizing the first term.

$\begin{aligned}&\frac{1}{1+\sqrt{2}}=\frac{1}{(1+\sqrt{2})} \times \frac{(1-\sqrt{2})}{(1-\sqrt{2})}=\frac{1-\sqrt{2}}{(1)^{2}-(\sqrt{2})^{2}} \\&=-(1-\sqrt{2})=\sqrt{2}-1 \\&\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=\frac{\sqrt{2}-\sqrt{3}}{2-3} \\&=\sqrt{3}-\sqrt{2}\end{aligned}$

Similarly, do for the second term

$\begin{aligned}&\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{\sqrt{3}-\sqrt{4}}{3-4}=-(\sqrt{3}-\sqrt{4})=\sqrt{4}-\sqrt{3} \\&\frac{1}{\sqrt{4}+\sqrt{5}}=\sqrt{5}-\sqrt{4} ; \frac{1}{\sqrt{5}+\sqrt{6}}=\sqrt{6}-\sqrt{5} \\&\frac{1}{\sqrt{6}+\sqrt{7}}=\sqrt{7}-\sqrt{6} ; \frac{1}{\sqrt{7}+\sqrt{8}}=\sqrt{8}-\sqrt{7} \\&\text { and } \frac{1}{\sqrt{8}+\sqrt{9}}=\sqrt{9}-\sqrt{8}\end{aligned}$

$\begin{aligned}&\text { L.H.S. }=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}} \\&+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}} \\&=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+(\sqrt{5}-\sqrt{4}) \\&+(\sqrt{6}-\sqrt{5})+(\sqrt{7}-\sqrt{6})+(\sqrt{8}-\sqrt{7})+(\sqrt{9}-\sqrt{8}) \\&=-1+\sqrt{9}=-1+3=2=\text { R.H.S. }\end{aligned}$

Therefore, the left hand side equates with right hand side, so the equation satisfies.

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