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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

N={0,$\mid x \mid $ ,$y$}

given: M=N

find :

$(x+\dfrac{1}{y})+(x^2+\dfrac{1}{y^2})+(x^3+\dfrac{1}{y^3})+--------+(x^{2001}+\dfrac{1}{y^{2001}})=?$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

N={0,$\mid x \mid $ ,$y$}

given: M=N

find :

$(x+\dfrac{1}{y})+(x^2+\dfrac{1}{y^2})+(x^3+\dfrac{1}{y^3})+--------+(x^{2001}+\dfrac{1}{y^{2001}})=?$

- Mar 31, 2013

- 1,322

x and y cannot be zero so we have xy = 1 as log(xy) = 0

N={0,$\mid x \mid $ ,$y$}

given: M=N

find :

$(x+\dfrac{1}{y})+(x^2+\dfrac{1}{y^2})+(x^3+\dfrac{1}{y^3})+--------+(x^{2001}+\dfrac{1}{y^{2001}})=?$

as x and y both are positive or -ve and we have |x| and y so both are > 0

so y = 1/x

so given sum

= 2 ( x + x^2 + ... + x^2001) = 2 x( 1- x^2001)/(1- x)

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- #3

- Jan 25, 2013

- 1,225

N={0,$\mid x \mid $ ,$y$}

given: M=N

find :

$(x+\dfrac{1}{y})+(x^2+\dfrac{1}{y^2})+(x^3+\dfrac{1}{y^3})+--------+(x^{2001}+\dfrac{1}{y^{2001}})=?$

- Mar 31, 2013

- 1,322

Now again as above xy =1 so m = { x, 1, 0}

N={0,$\mid x \mid $ ,$y$}

given: M=N

find :

$(x+\dfrac{1}{y})+(x^2+\dfrac{1}{y^2})+(x^3+\dfrac{1}{y^3})+--------+(x^{2001}+\dfrac{1}{y^{2001}})=?$

So |x| has to be 1 else if y = 1 and then x =1 and from N |x| cannot be y

So x = -1 , y = -1, and M = { -1, 1, 0} and N = { 0, 1 , - 1 }

So x = y =-1 and sum = x^n + 1/y^n = -2 for n = odd and 2 for n = even

So sum = -2 as 1st 1000 pairs cancel leaving with x^2001 + 1/y^2001 = - 2

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- #5

- Jan 25, 2013

- 1,225

kaliprasad :yes,very good solution