Playing with Square Root

Algebra Level 3

The sum

1 2 1 + 1 2 + 1 3 2 + 2 3 + 1 4 3 + 3 4 + + 1 100 99 + 99 100 \displaystyle \frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\dots+\frac{1}{100\sqrt{99}+99\sqrt{100}}

can be expressed as a b , \frac{a}{b}, where a a and b b are coprime positive integers.

What is the value of a + b a+b ?


This problem is adapted from a past year KMC Contest.


The answer is 19.

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4 solutions

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

= n = 1 99 1 n n + 1 + ( n + 1 ) n = \displaystyle \sum_{n=1}^{99} \frac{1}{n\sqrt{n+1} + (n+1)\sqrt{n}}

= n = 1 99 1 ( n ) ( n + 1 ) ( n + n + 1 ) = \displaystyle \sum_{n=1}^{99} \frac{1}{\sqrt{(n)(n+1)}(\sqrt{n} + \sqrt{n+1})}

= n = 1 99 1 ( n ) ( n + 1 ) ( n + n + 1 ) × n + 1 n n + 1 n = \displaystyle \sum_{n=1}^{99} \frac{1}{\sqrt{(n)(n+1)}(\sqrt{n} + \sqrt{n+1})} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}}

= n = 1 99 n + 1 n ( n ) ( n + 1 ) ( ( n + 1 ) 2 ( n ) 2 ) = \displaystyle \sum_{n=1}^{99} \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{(n)(n+1)}((\sqrt{n+1})^2 - (\sqrt{n})^2)}

= n = 1 99 n + 1 n ( n ) ( n + 1 ) = \displaystyle \sum_{n=1}^{99} \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{(n)(n+1)}}

= n = 1 99 1 n 1 n + 1 =\displaystyle \sum_{n=1}^{99} \sqrt{\frac{1}{n}} - \sqrt{\frac{1}{n+1}}

= ( 1 1 1 2 ) + ( 1 2 1 3 ) + + ( 1 99 1 100 ) = (\sqrt{\frac{1}{1}} - \sqrt{\frac{1}{2}}) + ( \sqrt{\frac{1}{2}} - \sqrt{\frac{1}{3}}) + \ldots + (\sqrt{\frac{1}{99}} - \sqrt{\frac{1}{100}})

= 1 1 1 100 = \sqrt{\frac{1}{1}} - \sqrt{\frac{1}{100}}

= 1 1 10 = 1 - \frac{1}{10}

= 9 10 = \frac{9}{10}

Therefore, a = 9 , b = 10 , a = 9, b = 10, and a + b = 19 a + b = \boxed{19}

super

Shreekant Khaitan - 7 years, 2 months ago

Only 4th step is wrong

archu hapse - 7 years, 2 months ago

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Sorry, Corrected now.(Hopefully).

Siddhartha Srivastava - 7 years, 2 months ago

Good One. By the way, did you use latex?

Ceesay Muhammed - 6 years, 6 months ago

what do you call the sign that looks like letter "E"?

JejeRem JereRem - 7 years, 2 months ago

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Sigma? Σ \Sigma <- This one?

Its used to signify summation.

http://en.wikipedia.org/wiki/%E2%88%91

Siddhartha Srivastava - 7 years, 2 months ago
Sharad Gaikwad
Mar 20, 2014

By rationalizing the denominator of each term we can get the series equal to 1- 1/sqrt(2)+ 1/sqrt(2)- 1/sqrt(3)+ 1/sqrt(3)- 1/sqrt(4) +1/sqrt(4)+...-1/sqrt(99)+1/sqrt(99)-1/sqrt(100) = 1-1/10= 9/10 and so a+b= 9+10=19!

This one's easier

Akshit Katiyar - 7 years, 2 months ago

{1/(2root1+1root2)}+{1/(3root2+2root3)}+{1/(4root3+3root4)}+------------+{1/(100root99+99root100)}=[1/root2root1(root2+root1)]+[1/root3root2(root3+root2)]+[1/root4root3(rot4+root3)]+-----------+[1/root100root99(root100+root99)]

after rationalizing we get

common terms cancels out and we left with

1-(1/root2)+(1/root2)-(1/root3)+(1/root3)-(1/root4)+(1/root4)-------------------(1/root99)+(1/root99)-(1/root100)=(1-1/1root100)=1-(1/10)=9/10=a/b.

so, a=9, b=10, a+b=9+10=19. Is the required answer.

Nice solution, but maybe you should use Latex next time.

Shashank Rammoorthy - 5 years, 11 months ago

rationalize each term and split them as single term we get sqrt1-1/sqrt2+1/sqrt2-1/sqrt3+...........+1/sqrt99-1/sqrt100 which is equal to 1-1/10=9/10. a+b=9+10=19

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