Question #8e330

2 Answers
Ratnaker Mehta
May 28, 2017

# 2.#

Explanation:

The Expression
#=1/(sqrt2+sqrt1)+1/(sqrt3+sqrt2)+1/(sqrt4+sqrt3)+...+1/(sqrt9+sqrt8),#

#=1/(sqrt2+sqrt1)xx(sqrt2-sqrt1)/(sqrt2-sqrt1)#
#+1/(sqrt3+sqrt2)xx(sqrt3-sqrt2)/(sqrt3-sqrt2)#
#+1/(sqrt4+sqrt3)xx(sqrt4-sqrt3)/(sqrt4-sqrt3)#
#+vdots #
#+1/(sqrt9+sqrt8)xx(sqrt9-sqrt8)/(sqrt9-sqrt8)#

#=(sqrt2-sqrt1)/(2-1)+(sqrt3-sqrt2)/(3-2)+...+(sqrt9-sqrt8)/(9-8)#

#=(cancelsqrt2-1)+(cancelsqrt3-cancelsqrt2)+(cancelsqrt4-cancelsqrt3)+...+(sqrt9-cancelsqrt8)#

#=3-1#

#=2.#

Enjoy Maths.!

Narad T.
May 28, 2017

The answer is #=2#

Explanation:

We need

#(a+b)(a-b)=a^2-b^2#

The general term is

#u_n=1/(sqrtn+sqrt(n+1))#

Simplifying the denominator

#u_n=1/(sqrt(n+1)+sqrt(n))*(sqrt(n+1)-sqrtn)/(sqrt(n+1)-sqrtn)#

#=(sqrt(n+1)-sqrtn)/(n+1-n)#

#=(sqrt(n+1)-sqrtn)#

Therefore,

#n=1#, #=>#, #u_1=cancelsqrt2-1#

#n=2#, #=>#, #u_2=cancelsqrt3-cancelsqrt2#

#n=3#, #=>#, #u_3=cancelsqrt4-cancelsqrt3#

#n=7#, #=>#, #u_7=cancelsqrt8-cancelsqrt7#

#n=8#, #=>#, #u_8=sqrt9-cancelsqrt8#

#sum_(n=1)^8u_n=sqrt9-1=3-1=2#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1160 views around the world
You can reuse this answer
Creative Commons License