Let r be a real number and n in N be such tha | vedclass

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(B) Given that $2x^2+2x+1$ divides $(x+1)^n-r$, the roots of $2x^2+2x+1=0$ must satisfy $(x+1)^n-r=0$.Solving $2x^2+2x+1=0$ using the quadratic formul

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$(4000, \frac{1}{4^{1000}})$

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(B) Given that $2x^2+2x+1$ divides $(x+1)^n-r$, the roots of $2x^2+2x+1=0$ must satisfy $(x+1)^n-r=0$.Solving $2x^2+2x+1=0$ using the quadratic formula:$x = \frac{-2 \pm \sqrt{4-8}}{4} = \frac{-1 \pm i}{2}$.Substituting $x = \frac{-1+i}{2}$ into $(x+1)^n = r$:$(\frac{-1+i}{2} + 1)^n = r \Rightarrow (\frac{1+i}{2})^n = r$.Expressing $\frac{1+i}{2}$ in polar form: $\frac{1+i}{2} = \frac{1}{\sqrt{2}} e^{i\pi/4}$.So, $(\frac{1}{\sqrt{2}} e^{i\pi/4})^n = r \Rightarrow \frac{1}{2^{n/2}} e^{in\pi/4} = r$.Since $r$ is a real number, the imaginary part must be zero, so $\sin(\frac{n\pi}{4}) = 0$, which implies $n$ must be a multiple of $4$.For $n=4000$, $r = \frac{1}{2^{4000/2}} \cos(\frac{4000\pi}{4}) = \frac{1}{2^{2000}} \cos(1000\pi) = \frac{1}{2^{2000}} = \frac{1}{4^{1000}}$.Thus, $(n, r) = (4000, \frac{1}{4^{1000}})$.

Let $r$ be a real number and $n \in N$ be such that the polynomial $2x^2+2x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be

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