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Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^{2}+2 x+1$ divides the polynomial $(x+1)^{n}-r$. Then $(\mathrm{n}, \mathrm{r})$ can be-
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The correct answer is:
$\left(4000, \frac{1}{4^{1000}}\right)$
$$
\begin{array}{l}
2 x^{2}+2 x+1=0 \\
x=\frac{-1+i}{2}, \frac{-1-i}{2} \\
x \text { satisfies }(x+1)^{n}-r=0 \\
\left(\frac{-1 \pm i}{2}+1\right)^{n}-r=0 \\
\left(\frac{1 \pm i}{2}\right)^{n}-r=0 \\
\left(\frac{1}{\sqrt{2}}\right)^{n}\left(\frac{1+i}{\sqrt{2}}\right)^{n}=r \\
\left(\frac{1}{\sqrt{2}}\right)^{n}\left(e^{\pm \frac{i \pi}{4}}\right)^{n}=r \\
R H S=\text { real } \\
\text { LHS }=\text { real only when } n=\text { multiply of } 4 \\
n=4000 \\
r=\left(\frac{1}{\sqrt{2}}\right)^{4000}=\frac{1}{4^{1000}}
\end{array}
$$
\begin{array}{l}
2 x^{2}+2 x+1=0 \\
x=\frac{-1+i}{2}, \frac{-1-i}{2} \\
x \text { satisfies }(x+1)^{n}-r=0 \\
\left(\frac{-1 \pm i}{2}+1\right)^{n}-r=0 \\
\left(\frac{1 \pm i}{2}\right)^{n}-r=0 \\
\left(\frac{1}{\sqrt{2}}\right)^{n}\left(\frac{1+i}{\sqrt{2}}\right)^{n}=r \\
\left(\frac{1}{\sqrt{2}}\right)^{n}\left(e^{\pm \frac{i \pi}{4}}\right)^{n}=r \\
R H S=\text { real } \\
\text { LHS }=\text { real only when } n=\text { multiply of } 4 \\
n=4000 \\
r=\left(\frac{1}{\sqrt{2}}\right)^{4000}=\frac{1}{4^{1000}}
\end{array}
$$
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