Question

Challenger question
Accept challenge
(question taken from aakash book)

Answer 1

Solution :-

Q: 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

L.H.S, 1/(1 + √2) × (1 - √2)/(1 - √2) + 1/(√2 + √3) × (√2 - √3)/(√2 - √3) + 1/(√3 + √4) × (√3 - √4)/(√3 - √4) + 1/(√4 + √5) × (√4 - √5)/(√4 - √5) + 1/(√5 + √6) × (√5 - √6)/(√5 - √6) + 1/(√6 + √7) × (√6 - √7)/(√6 - √7) + 1/(√7 + √8) × (√7 - √8)/(√7 - √8) + 1/(√8 + √9) × (√8 - √9)/(√8 - √9)

= (1 - √2)/(1 - 2) + (√2 - √3)/(2 - 3) + (√3 - √4)/(3 - 4) + (√4 - √5)/(4 - 5) + (√5 - √6)/(5 - 6) + (√6 - √7)/(6 - 7) + (√7 - √8)/(7 - 8) + (√8 - √9)/(8 - 9)

= (1 - √2)/(-1) + (√2 - √3)/(-1) + (√3 - √4)/(-1) + (√4 - √5)/(-1) + (√5 - √6)/(-1) + (√6 - √7)/(-1) + (√7 - √8)/(-1) + (√8 - √9)/(-1)

= - (1 - √2) - (√2 - √3) - (√3 - √4) - (√4 - √5) - (√5 - √6) - (√6 - √7) - (√7 - √8) - (√8 - √9)

= - 1 + √2 - √2 + √3 - √3 + √4 - √4 + √5 - √5 + √6 - √6 + √7 - √7 + √8 - √8 + √9

= - 1 + √9

= - 1 + 3

= 2 _______ R.H.S

Hence, proved

Answer 2

$\huge \underline \mathfrak {Solution:-}$

LHS =

$\frac{1}{ 1 + \sqrt{2} } + \frac{1}{ \sqrt{ 2} + \sqrt{3} } + \frac{1}{ \sqrt{3 } + \sqrt{4} } \\ + \frac{1}{ \sqrt{ 4} + \sqrt{5} } + \frac{1}{ \sqrt{ 5} + \sqrt{6} } + \frac{1}{ \sqrt{ 6} + \sqrt{7} } \\ + \frac{1}{ \sqrt{ 7} + \sqrt{8} } + \frac{1}{ \sqrt{8 } + \sqrt{9} } \\ \\ on \: rationalising \: them\\ \\ = \frac{1 - \sqrt{2} }{1 - 2} + \frac{ \sqrt{2} - \sqrt{3} }{2 - 3} + \frac{ \sqrt{3} - \sqrt{4} }{3 - 4} \\ \\ + \frac{ \sqrt{4} - \sqrt{5} }{4 - 5} + \frac{ \sqrt{5} - \sqrt{6} }{5 - 6} + \frac{ \sqrt{6} - \sqrt{7} }{6 - 7} \\ \\ + \frac{ \sqrt{7} - \sqrt{8} }{7 - 8} + \frac{ \sqrt{8} - \sqrt{9} }{8 - 9}$

$= - 1 + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} \\ \\ + \sqrt{4} - \sqrt{4} + \sqrt{5} - \sqrt{5} + \sqrt{6} - \sqrt{6} \\ \\ + \sqrt{7} - \sqrt{7} + \sqrt{8} - \sqrt{8} + \sqrt{9} \\ \\ = - 1 + \sqrt{9} \\ \\ = - 1 + \sqrt{3 \times 3} \\ \\ = - 1 + 3 \\ \\ = 2$

=RHS

Hence proved.