Problem 13

Calculus Level 4

Let a n { a }_{ n } be a sequence defined on positive integers as a n = 4 n + 4 n 2 − 1 2 n + 1 + 2 n − 1 { a }_{ n }=\frac { 4n+\sqrt { { 4n }^{ 2 }-1 } }{ \sqrt { 2n+1 } +\sqrt { 2n-1 } } Find the value of a 1 + a 2 + . . . . . . . . + a 60 \\ { a }_{ 1 }+{ a }_{ 2 }+........+{ a }_{ 60 }


The answer is 665.

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Utkarsh Bansal by Utkarsh Bansal , 23, India

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1 solution

Patrick Corn
Mar 6, 2015

Write x = 2 n + 1 , y = 2 n − 1 x = \sqrt{2n+1}, y = \sqrt{2n-1} . Then a n = x 2 + y 2 + x y x + y = x 3 − y 3 x 2 − y 2 = x 3 − y 3 2 . a_n = \frac{x^2+y^2+xy}{x+y} = \frac{x^3-y^3}{x^2-y^2} = \frac{x^3-y^3}2.

So ∑ n = 1 60 a n = 1 2 ∑ n = 1 60 ( ( 2 n + 1 ) 3 / 2 − ( 2 n − 1 ) 3 / 2 ) \sum_{n=1}^{60} a_n = \frac12 \sum_{n=1}^{60} \left( (2n+1)^{3/2} - (2n-1)^{3/2} \right) which telescopes to 1 2 ( 121 ) 3 / 2 − 1 2 ( 1 ) 3 / 2 = 665 \frac12(121)^{3/2} - \frac12(1)^{3/2} = \fbox{665} .

10 Helpful 1 Interesting 0 Brilliant 0 Confused

Nice solution Patrick sir , upvoted

Utkarsh Bansal - 6 years, 3 months ago

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