IIT JEE , AIEEE Forum - Mathematics , Algebra

Radon222

$Question: \text{If } x+\frac{1}{x}=1 \text{ then find } x^{4000}+\frac{1}{x^{4000}}$

I used two different methods and i am getting different answers from both the methods, pls explain why?

Method 1 :

$x+\frac{1}{x}=1 \\ x^2-x+1=0 \\ x=-\omega,-\omega^2 \\ \rightarrow x^{4000}+\frac{1}{x^{4000}}=\omega^{4000}+\frac{1}{\omega^{4000}}=\frac{ \omega ^2+1}{\omega} \\ =\frac{-\omega}{\omega}=-1$

Method 2:

squaring both sides again and again of $x+\frac{1}{x}=1$ we get a general result

$x^{2^n}+\frac{1}{x^{2^n}}=1-2n$

now putting n=11 we get
$x^{2048}+\frac{1}{x^{2048}} = -21 \\ \text{ and by putting n=12 we get } x^{4096}+\frac{1}{x^{4096}} =-23 \\ \text{this means } x^{4000}+\frac{1}{x^{4000}} \text{ should be around -23 and not -1 as we get from method 1}$

please explain why is the answer coming different in the two cases?

ankur.kkhurana

method one is right and for method two the general result u arrived is wrong
it is not 1-2n
but on right hand side it is 1-2 =-1 hence answer is -1

Radon222

but ankur i find that the general result is true and is being obeyed for n=0,1,2,3.......

arpan1

how is it -1 on the right side

arpan1

@radon

what values of x did u put in the second method

computer001

radon...in 2nd method x^(2^n)+ 1/x^(2^n)= -1 always ie it come
(1-2) not 1-2n pl chkup

Radon222

sorry that was a real silly mistake i got my doubt cleared myself....I like a fool was not squaring right hand side and kept squaring left hand side...sorry to waste everyone's time(and mine too)

babanbanerjee

The first method looks fine.
In the 2nd method
squaring u will get
(x^2n+1/x^2n+2=1)[(a+b)^2=a^2+b^2+2ab]
So ur general result in the end comes out to be wrong.

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