If binom{n-1}{r} = (k^2 - 3) binom{n}{r+1},th | vedclass

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(D) Given the equation $\binom{n-1}{r} = (k^2 - 3) \binom{n}{r+1}$.We know that $\binom{n}{r+1} = \frac{n}{r+1} \binom{n-1}{r}$.Substituting this into

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$[-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$

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(D) Given the equation $\binom{n-1}{r} = (k^2 - 3) \binom{n}{r+1}$.We know that $\binom{n}{r+1} = \frac{n}{r+1} \binom{n-1}{r}$.Substituting this into the equation,we get $\binom{n-1}{r} = (k^2 - 3) \frac{n}{r+1} \binom{n-1}{r}$.Assuming $\binom{n-1}{r} 
eq 0$,we have $1 = (k^2 - 3) \frac{n}{r+1}$,which implies $k^2 - 3 = \frac{r+1}{n}$.Since $0 \leq r \leq n-1$,we have $1 \leq r+1 \leq n$.Dividing by $n$,we get $\frac{1}{n} \leq \frac{r+1}{n} \leq 1$.As $n$ can be any positive integer,the range of $\frac{r+1}{n}$ is $(0, 1]$.Thus,$0 Solving for $k$,we get $k \in [-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$.

If $\binom{n-1}{r} = (k^2 - 3) \binom{n}{r+1}$,then $k \in \dots$

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