What is the value of 1/(1 + 2^(1/2)) + 1/(2^(1/2) + 3^(1/2)) + 1/(2016

Question

Competition Mode Question

What is the value of  \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{2016} + \sqrt{2017}}  ?

A.  \sqrt{2016} - 1

B.  \sqrt{2017} -1

C.  \sqrt{2017}

D.  \sqrt{2016} + 1

E.  \sqrt{2017} + 1

unraveled · Accepted Answer

What is the value of  \frac{1}{1+√2} + \frac{1}{√2+√3} + \frac{1}{√3+√4} + ...+\frac{1}{√2016+√2017} ?

A. √12016−1

B. √2017−1

C. √2017

D. √2016+1

E. √2017+1

\frac{1}{1+√2} + \frac{1}{√2+√3} + \frac{1}{√3+√4} + ...+\frac{1}{√2016+√2017} 
 \frac{1}{1+√2}*\frac{1-√2}{1-√2} + \frac{1}{√2+√3}*\frac{√2-√3}{√2-√3} + \frac{1}{√3+√4}*\frac{√3-√4}{√3-√4} + ...+\frac{1}{√2016+√2017}*\frac{√2016-√2017}{√2016-√2017} 
 \frac{1-√2}{1-2} + \frac{√2-√3}{2-3} + \frac{√3-√4}{3-4} + ...+\frac{√2016-√2017}{2016-2017} 
-1 + √2 - √2 + √3 - √3 + √4 - √4 ... + √2016 - √2016 + √2017
√2017 - 1

Answer B.
  ]

Shrey08 · Answer

If we rationalize the denominator of each fraction in the series we get:

\frac{(1 - \sqrt{2})}{(1^2-(\sqrt{2})^2)} + \frac{(\sqrt{2} - \sqrt{3})}{((\sqrt{2})^2-(\sqrt{3})^2)} + \frac{(\sqrt{3} - \sqrt{4})}{((\sqrt{3})^2-(\sqrt{4})^2)}..............+\frac{(\sqrt{2016} - \sqrt{2017})}{((\sqrt{2016})^2-(\sqrt{2017})^2)}

in this series denominator of every fraction is  -1

so the series becomes

-(1 - \sqrt{2})-(\sqrt{2} - \sqrt{3}) -(\sqrt{3} - \sqrt{4}).........-(\sqrt{2016} - \sqrt{2017})

= -1 + \sqrt{2} - \sqrt{2} + \sqrt{3}- \sqrt{3} + \sqrt{4}........ - \sqrt{2016} + \sqrt{2017}
 
after cancelling out the pairs with opposite signs what remains is:
 -1 + \sqrt{2017}

\sqrt{2017}-1

Option B

ArunSharma12 · Answer

\frac{1}{1+√2} = \frac{1 - √2}{(1+√2)(1-√2)} = \frac{1 - √2}{1-2} = √2 - 1

\frac{1}{1+√2}+\frac{1}{√2+√3}+\frac{1}{√3+√4}+...+\frac{1}{√2016+√2017}  =

√2-1 + √3 - √2 + √4 - √3 +...+ √2017 - √2016  =  √2017 - 1 
ans: B

Lipun · Answer

Let's start with the second term.

\frac{1}{(\sqrt{2}+\sqrt{3})}

=  \frac{3-2}{(\sqrt{2}+\sqrt{3})} ,

now apply  a^2 - b^2 = (a+b)(a-b)

=  \frac{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}{(\sqrt{2}+\sqrt{3})}

=  (\sqrt{3}-\sqrt{2})

After simplifying all the terms in the expression, it reduces to  \sqrt{2017} - 1

Ans: B

exc4libur · Answer

rationalize the denomintor:
1/(1 + 2^(1/2))*(1 - 2^(1/2))/(1 - 2^(1/2)),
(1 - 2^(1/2))/(1^2 - 2^(1/2)^2)=(1 - 2^(1/2))/-1
1/(2^(1/2) + 3^(1/2))*(2^(1/2) - 3^(1/2))/(2^(1/2) - 3^(1/2)),
(2^(1/2) - 3^(1/2))/2-3=(2^(1/2) - 3^(1/2))/-1
....
(2016^(1/2) - 2017^(1/2))/-1

they all have the same denominator (-1):
  / -1
=   / -1
= -1 + 2017^(1/2)

Ans (B)

rajatchopra1994 · Answer

Here genral term Tn can br represented as

Tn= 1/(√(n)+√(n+1))

Now multiplying and dividing every term by

√(n+1)-√(n) we get

Tn=  /(n+1-n)

Tn=  /1

Now we observe

S= √(2)- √(1) + √(3)- √(2)+ ...... +√(n+1)- √(n)+…... √(2017)- √(2016)

In this above series, we are left with only

S= -v(1)+ √(2017)

S= √2017 – 1

IMO-B

felipet190 · Answer

Bunuel can you please unhide my post that you incorrectly deleted? I know the question I posted was discussed in a separate post. I was not seeking to discuss the question - I was seeking to discuss the solution provided by the authors of my book. Please and thank you. gmatclub /forum/product-of-two-integers-with-units-digit-6-has-unit-digit-4-no-327186.html#p2548781

monikakumar · Answer

\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\frac{1}{\sqrt{3}+\sqrt{4}}

taking conjugates,

\frac{\sqrt{2}-1}{4-1}\frac{\sqrt{4}-\sqrt{3}}{16-9}

so,
the denominator has odd numbers like 3,5,7,9,11,13..
taking LCM cancels out the denominator.

Hence,

\sqrt{2017}-1

Ans B

neenubabu · Answer

1/(1+√2) + 1/(√2+√3) + 1/(√3+√4)+…………….+1/(√2016+√2017)
Multiply the numerator and denominator with a-b where higher number in the denominator is considered as ‘a’ and lower as ‘b’
Expression will be as follows
(√2-1) /(1+√2) (√2-1) + (√3-√2) /(√2+√3) (√3-√2) 
+ (√4-√3) /(√3+√4) (√4-√3) +…………….+(√2017-√2016) /(√2016+√2017)(√2017-√2016)

Denominator is of the form (a+b)(a-b) = a^2 -b^2
Here are in all the denominator a^2-b^2 = 1
Then the expression becomes as follows

(√2-1) +(√3-√2)+ (√4-√3)+……+ (√2017-√2016) = √2017-1
IMO Option B is the correct answer

Bunuel · Answer

You should not re-post existing topic. Any additional question should be posted in existing thread.

bumpbot · Answer

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What is the value of 1/(1 + 2^(1/2)) + 1/(2^(1/2) + 3^(1/2)) + 1/(2016

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