SOLUTION: Find the sum of the series 5/(sqrt2+sqrt1) + 5/(sqrt3+sqrt2) + 5/(sqrt4+sqrt3) + 5/(sqrt5+sqrt4)+...+ 5/(sqrt36)+(sqrt35)

Algebra ->  Sequences-and-series -> SOLUTION: Find the sum of the series 5/(sqrt2+sqrt1) + 5/(sqrt3+sqrt2) + 5/(sqrt4+sqrt3) + 5/(sqrt5+sqrt4)+...+ 5/(sqrt36)+(sqrt35)      Log On


   

Question 1149797: Find the sum of the series 5/(sqrt2+sqrt1) + 5/(sqrt3+sqrt2) + 5/(sqrt4+sqrt3) + 5/(sqrt5+sqrt4)+...+ 5/(sqrt36)+(sqrt35)
Answer by ikleyn(52864) About Me  (Show Source):
You can put this solution on YOUR website!
.

HINT.  Multiply each denominator (and numerator) by the conjugate number to the denominator.


In each denominator, you will get the value of 1.


The numerators will form alternated sequence.


As a result, all intermediate terms will cancel each other, and you will get the combination 
of the last and the first terms as your answer.


ANSWER.  5*sqrt(36) - 5 = 5*6 - 5 = 25.


See the lesson
    - Amazing calculations with fractions that contain quadratic irrationalities in denominators
in this site, Problem 4.

Your problem is solved there with all detailed explanations.