^{n-1}C_r = (k^2 - 8) ^nC_{r+1} if and only i | vedclass

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(A) Given the equation: $^{n-1}C_r = (k^2 - 8) ^nC_{r+1}$Using the identity $^{n}C_{r+1} = \frac{n}{r+1} ^{n-1}C_r$,we can write:$^{n-1}C_r = (k^2 - 8

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$2\sqrt{2} < k \leq 3$

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(A) Given the equation: $^{n-1}C_r = (k^2 - 8) ^nC_{r+1}$Using the identity $^{n}C_{r+1} = \frac{n}{r+1} ^{n-1}C_r$,we can write:$^{n-1}C_r = (k^2 - 8) \cdot \frac{n}{r+1} ^{n-1}C_r$Since $^{n-1}C_r 
eq 0$,we have:$1 = (k^2 - 8) \frac{n}{r+1} \Rightarrow \frac{r+1}{n} = k^2 - 8$Since $0 \leq r+1 \leq n$,it follows that $0  0$ for non-trivial combinations).Thus,$0 Solving $k^2 - 8 > 0$ gives $k^2 > 8$,so $k > 2\sqrt{2}$ or $k Solving $k^2 - 8 \leq 1$ gives $k^2 \leq 9$,so $-3 \leq k \leq 3$.Combining these,for positive $k$,we get $2\sqrt{2} < k \leq 3$.

$^{n-1}C_r = (k^2 - 8) ^nC_{r+1}$ if and only if:

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