Question

(1/√2+1)+(1/√3+√2)+(1/√4+√3a)+...........+(1/√a+√8)

Answer 1

Answer:

$=\frac{1}{\sqrt2+\sqrt1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}.................+\frac{1}{\sqrt9+\sqrt8}=2$

Step-by-step explanation:

Given;

$\frac{1}{\sqrt2+\sqrt1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}.................+\frac{1}{\sqrt{a}+\sqrt8}$

By observation, each term is of form

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

Then, clearly a=8+1=9

$\frac{1}{\sqrt2+\sqrt1}$

$=\frac{1}{\sqrt2+\sqrt1}.\frac{\sqrt2-\sqrt1}{\sqrt2-\sqrt1}$

$=\frac{\sqrt2-\sqrt1}{(\sqrt2)^2-(\sqrt1)^2}$

$=\frac{\sqrt2-\sqrt1}{2-1}$

$=\frac{\sqrt2-\sqrt1}{1}$

$=\sqrt2-\sqrt1$

This implies,

$\frac{1}{\sqrt2+\sqrt1}=\sqrt2-\sqrt1$

similarly,

$\frac{1}{\sqrt3+\sqrt2}=\sqrt3-\sqrt2$

$\frac{1}{\sqrt4+\sqrt3}=\sqrt4-\sqrt3$

.

.

.

$\frac{1}{\sqrt9+\sqrt8}=\sqrt9-\sqrt8$

Now,

$\frac{1}{\sqrt2+\sqrt1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}.................+\frac{1}{\sqrta+\sqrt8}$

$=\frac{1}{\sqrt2+\sqrt1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}.................+\frac{1}{\sqrt9+\sqrt8}$

$=(\sqrt2-\sqrt1)+(\sqrt3-\sqrt2)+(\sqrt4-\sqrt3)+..................+(\sqrt8-\sqrt7)+(\sqrt9-\sqrt8)$

$=(\sqrt2-\sqrt1)+(\sqrt3-\sqrt2)+(\sqrt4-\sqrt3)+..................+(\sqrt8-\sqrt7)+(\sqrt9-\sqrt8)$

$=\sqrt9-\sqrt1$

$=3-1$

$=2$