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1/7 -1/11 is it + or - between these two terms?

Sboosy is right !

will u consider giving solutions ?

Wrong !

The coefficients of x^2y^2

, $yzt^2 \text{ and } xyzt$ in the expansion of (x+y+z+t)⁴are in what ratio?

Phew ! atleast my nickels will not go waste now

If $\displaystyle\sum_{r=0}^{n}\left[ \frac {r+2}{r+1} \right] .^nC_r=\frac{2^8-1}{6}$ ,then n is equal to ??

wow ! great attempts ....and really awesome substitution by konichiwa

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)

that was not worth to be called "the demon"

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

rahul your answer is perfect except a small negative sign mistake in the end

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

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?