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If the third term in the binomial expansion of $(1+x)^m$ is $-\frac{1}{8} x^2$, then the rational value of $m$ is
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The correct answer is:
$1/2$
We have $(1+x)^m=1+m x+\frac{m(m-1)}{2!} x^2+\ldots$
By hypothesis, $\frac{m(m-1)}{2} x^2=-\frac{1}{8} x^2$
$\Rightarrow 4 m^2-4 m=-1 \Rightarrow(2 m-1)^2=0 \Rightarrow m=\frac{1}{2}$
By hypothesis, $\frac{m(m-1)}{2} x^2=-\frac{1}{8} x^2$
$\Rightarrow 4 m^2-4 m=-1 \Rightarrow(2 m-1)^2=0 \Rightarrow m=\frac{1}{2}$
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