Question

What will be the value of 1/2+√7+1/√7+√10+....+1/√28+√31 .plzz its very urgent guys.going to appear the aakash exams plzzzz

Answer 1

Answer-

$\frac{1}{2+\sqrt{7}} +\frac{1}{\sqrt{7}+\sqrt{10}}+......+\frac{1}{\sqrt{28}+\sqrt{31}}=\frac{\sqrt{31}-2}{3}$

Solution-

Given,

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

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

Multiplying conjugates in each terms,

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

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

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

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

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

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

Only √4 from the first fraction and √31 from the last fraction will remain as it is, and all the other terms will cancelled out each other. So it becomes

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

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

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

$\therefore \frac{1}{2+\sqrt{7}} +\frac{1}{\sqrt{7}+\sqrt{10}}+......+\frac{1}{\sqrt{28}+\sqrt{31}}=\frac{\sqrt{31}-2}{3}$