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Algebra Level 4

Find the value of n = 0 1000 ( x 2 n + y 2 n ) \prod _{ n=0}^{ 1000 }{ \left( { x }^{ { 2 }^{ n } }+{ y }^{ { 2 }^{ n } } \right) }

x 2 1001 y 2 1001 x y \frac { { x }^{ { 2 }^{ 1001 } }-{ y }^{ { 2 }^{ 1001 } } }{ x-y } x 2 1001 + y 2 1001 x + y \frac { { x }^{ { 2 }^{ 1001 } }+{ y }^{ { 2 }^{ 1001 } } }{ x+y } x 2 1001 y 2 1001 x + y \frac { { x }^{ { 2 }^{ 1001 } }-{ y }^{ { 2 }^{ 1001 } } }{ x+y } x 2 1001 + y 2 1001 x y \frac { { x }^{ { 2 }^{ 1001 } }+{ y }^{ { 2 }^{ 1001 } } }{ x-y }

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1 solution

Archit Boobna
Nov 5, 2014

Multiply it by x y x y \frac { x-y }{ x-y } and solve now.

x y x y ( x + y ) ( x 2 + y 2 ) . . . . . . . . . \frac { x-y }{ x-y } (x+y)({ x }^{ 2 }+{ y }^{ 2 }).........

= x 2 y 2 x y ( x 2 + y 2 ) . . . . . . . . . =\frac { { x }^{ 2 }-{ y }^{ 2 } }{ x-y } ({ x }^{ 2 }+{ y }^{ 2 })......... and so on.......

Hey, it is beginning with n=1 not n=0. Hence you should multiply by:- x 2 y 2 x 2 y 2 \dfrac{x^{2} -y^{2}}{x^{2}-y^{2}} Hence your answer would have been correct if you have begun it from n=0.

Prakhar Gupta - 6 years, 7 months ago

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thanks i hav edited it

Archit Boobna - 6 years, 7 months ago

This is valid iff x y x\neq y . I think you should add this condition in your problem (although it might be obvious).

By the way, here 's a similar problem. Take a look at the generalization mentioned in the comments in Sualeh Asif's solution there.

Prasun Biswas - 5 years, 11 months ago

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