Question

the value of 1/2+root7 + 1/root7+root10 +......+1/root28+root31 is

Answer 1

Rationalize the denominator
multiply and divide by √7-2

1/(√7 +2)
= (√7-2)/(√7+2)*(√7-2 )
= (√7-2)/(7-4 )
= (√7-2)/3

similarly terms will be
(√10-√7)/3
.
.
.
.

(√31-√28)/3

add all to get
(√31-2)/3
option 4 is the correct ans

Answer 2

Answer-

$D. \frac{\sqrt{31}-2}{3}$

Solution-

The given expression is,

$=\frac{1}{2+\sqrt{7}} +\frac{1}{\sqrt{7}+\sqrt{10}}+......+\frac{1}{\sqrt{28}+\sqrt{31}}$

$=\frac{1}{\sqrt{4}+\sqrt{7}} +\frac{1}{\sqrt{7}+\sqrt{10}}+......+\frac{1}{\sqrt{28}+\sqrt{31}}$

$=\frac{(\sqrt{4}-\sqrt{7})}{(\sqrt{4}+\sqrt{7})(\sqrt{4}-\sqrt{7})} +\frac{(\sqrt{7}-\sqrt{10})}{(\sqrt{7}+\sqrt{10})(\sqrt{7}-\sqrt{10})}+......+\frac{(\sqrt{28}-\sqrt{31})}{(\sqrt{28}+\sqrt{31})(\sqrt{28}-\sqrt{31})}$

$=\frac{(\sqrt{4}-\sqrt{7})}{(\sqrt{4})^2-(\sqrt{7})^2} +\frac{(\sqrt{7}-\sqrt{10})}{(\sqrt{7})^2-(\sqrt{10})^2}+......+\frac{(\sqrt{28}-\sqrt{31})}{(\sqrt{28})^2-(\sqrt{31})^2}$

$=\frac{(\sqrt{4}-\sqrt{7})}{4-7} +\frac{(\sqrt{7}-\sqrt{10})}{7-10}+......+\frac{(\sqrt{28}-\sqrt{31})}{28-31}$

$=\frac{(\sqrt{4}-\sqrt{7})}{-3} +\frac{(\sqrt{7}-\sqrt{10})}{-3}+......+\frac{(\sqrt{28}-\sqrt{31})}{-3}$

$=\frac{(\sqrt{7}-\sqrt{4})}{3} +\frac{(\sqrt{10}-\sqrt{7})}{3}+......+\frac{(\sqrt{31}-\sqrt{28})}{3}$

$= \frac{1}{3}[\sqrt{7}-\sqrt{4}}+\sqrt{10}-\sqrt{7}+......+\sqrt{31}-\sqrt{28}]$

One term from the first fraction (i.e √4) and other from the last (i.e √31) will remain eventually, and all the other terms will be cancelled out.

$= \frac{1}{3}[\sqrt{31}-\sqrt{4}]$

$= \frac{\sqrt{31}-\sqrt{4}}{3}$

$= \frac{\sqrt{31}-2}{3}$