Such A Long sequence

Algebra Level 4

1 3 1 + 3 + 1 5 3 + 3 5 . . . . . . . . . . . + 1 225 223 + 223 225 = ? \large \frac{1}{3\sqrt1+\sqrt3}+\frac{1}{5\sqrt3+3\sqrt5}...........+\frac{1}{225\sqrt{223}+223\sqrt{225}} = ?

If this sum can be represented as a b \frac{a}{b} , where both a , b a,b are square free and positive integers, find ( a $ b ) \large(a\$b) if the operation ( a $ b ) \large(a\$b) is defined as b ( m o d a ) \large b \pmod{a} .


The answer is 1.

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Krishna Ar by Krishna Ar , 21, India

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2 solutions

Arif Ahmed
Sep 19, 2014

∑ k = 1 112 1 ( 2 k + 1 ) 2 k − 1 + ( 2 k − 1 ) 2 k + 1 \sum _{ k = 1 }^{ 112 }{ \frac { 1 }{ (2k+1)\sqrt { 2k-1 } \quad +\quad (2k-1)\sqrt { 2k+1 } } }

0.5 ∑ k = 1 112 ( 2 k + 1 ) − ( 2 k − 1 ) ( 2 k + 1 ) 2 k − 1 + ( 2 k − 1 ) 2 k + 1 0.5\sum _{ k = 1 }^{ 112 }{ \frac { (2k + 1) - (2k - 1) }{ (2k+1)\sqrt { 2k - 1 } + (2k-1)\sqrt { 2k+1 } } }

0.5 ∑ k = 1 112 ( 2 k + 1 − 2 k − 1 ) ( 2 k + 1 + 2 k − 1 ) ( 2 k − 1 2 k + 1 ) × ( 2 k + 1 + 2 k − 1 ) 0.5\sum _{ k = 1 }^{ 112 }{ \frac { (\sqrt { 2k+1 } - \sqrt { 2k-1 } )(\sqrt { 2k+1 } +\sqrt { 2k-1 } ) }{ (\sqrt { 2k-1 } \sqrt { 2k+1 } )\times(\sqrt { 2k+1 } +\sqrt { 2k-1 } ) } }

0.5 ∑ k = 1 112 1 2 k − 1 − 1 2 k + 1 0.5\sum _{ k = 1 }^{ 112 }{ \frac { 1 }{ \sqrt { 2k-1 } } } -\frac { 1 }{ \sqrt { 2k+1 } }

This telescopes to 7 15 \frac { 7 }{ 15 }

Therefore answer is : 15 mod 7 = 1

8 Helpful 0 Interesting 0 Brilliant 0 Confused

how is 15mod7=1

Somnath KVS - 6 years, 5 months ago

Yes, Rationalizing will help.

∑ k = 1 112 1 ( 2 k + 1 ) 2 k − 1 + ( 2 k − 1 ) 2 k + 1 \displaystyle \sum\limits_{k=1}^{112} \frac{1}{(2k+1)\sqrt{2k-1} + (2k-1)\sqrt{2k+1}}

= ∑ k = 1 112 1 ( ( 2 k + 1 ) ( 2 k − 1 ) ) ( 2 k + 1 + 2 k − 1 ) = \displaystyle \sum\limits_{k=1}^{112} \frac{1}{(\sqrt{(2k+1)(2k-1)})(\sqrt{2k+1} + \sqrt{2k-1})}

= ∑ k = 1 112 2 k + 1 − 2 k − 1 ( ( 2 k + 1 ) ( 2 k − 1 ) ) ( 2 k + 1 + 2 k − 1 ) ( 2 k + 1 − 2 k − 1 ) = \displaystyle \sum\limits_{k=1}^{112} \frac{\sqrt{2k+1}-\sqrt{2k-1}}{(\sqrt{(2k+1)(2k-1)})(\sqrt{2k+1} + \sqrt{2k-1})(\sqrt{2k+1}-\sqrt{2k-1)}}

= ∑ k = 1 112 2 k + 1 − 2 k − 1 ( ( 2 k + 1 ) ( 2 k − 1 ) ) × 2 = \displaystyle \sum\limits_{k=1}^{112} \frac{\sqrt{2k+1}-\sqrt{2k-1}}{(\sqrt{(2k+1)(2k-1)})\times 2}

= 1 2 × ∑ k = 1 112 ( 1 2 k − 1 − 1 2 k + 1 ) = \displaystyle \frac{1}{2} \times \sum\limits_{k=1}^{112} \left(\frac{1}{\sqrt{2k-1}} - \frac{1}{\sqrt{2k+1}}\right)

= 1 2 × ( 1 − 1 15 ) = \displaystyle \frac{1}{2}\times \left(1 - \frac{1}{15}\right)

= 7 15 = \displaystyle \boxed{\frac{7}{15}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice title! :D

milind prabhu - 6 years, 7 months ago

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