Question

Crn−1=(k2−8)Cr+1n if and only if :

Answer 1

k^2-8 = \frac{r(r+1)}{(n-r)(n-r+1)}

Answer 2

To solve the equation $C_{r-1}^n = (k^2-8)C_{r+1}^n$, we will use the definition of combinations, $C_x^n = \frac{n!}{x!(n-x)!}$.

First, let's write out the terms for $C_{r-1}^n$ and $C_{r+1}^n$: $C_{r-1}^n = \frac{n!}{(r-1)!(n-(r-1))!} = \frac{n!}{(r-1)!(n-r+1)!}$

$C_{r+1}^n = \frac{n!}{(r+1)!(n-(r+1))!} = \frac{n!}{(r+1)!(n-r-1)!}$

Now, substitute these expressions into the given equation: $\frac{n!}{(r-1)!(n-r+1)!} = (k^2-8) \frac{n!}{(r+1)!(n-r-1)!}$

Assuming $n! \neq 0$ (which is true for $n \ge 0$), we can cancel $n!$ from both sides: $\frac{1}{(r-1)!(n-r+1)!} = (k^2-8) \frac{1}{(r+1)!(n-r-1)!}$

To isolate $(k^2-8)$, multiply both sides by $(r+1)!(n-r-1)!$: $k^2-8 = \frac{(r+1)!(n-r-1)!}{(r-1)!(n-r+1)!}$

Next, we expand the factorials in the numerator and denominator to find common terms that can be cancelled: $(r+1)! = (r+1) \cdot r \cdot (r-1)!$ $(n-r+1)! = (n-r+1) \cdot (n-r) \cdot (n-r-1)!$

Substitute these expanded forms back into the equation for $(k^2-8)$: $k^2-8 = \frac{(r+1) \cdot r \cdot (r-1)! \cdot (n-r-1)!}{(r-1)! \cdot (n-r+1) \cdot (n-r) \cdot (n-r-1)!}$

Now, cancel out the common factorial terms, $(r-1)!$ and $(n-r-1)!$: $k^2-8 = \frac{r(r+1)}{(n-r)(n-r+1)}$

This is the condition "if and only if" the given equation holds. For the combinations to be defined, we must have $r-1 \ge 0 \implies r \ge 1$ and $n \ge r+1$. Under these conditions, all terms in the expression $\frac{r(r+1)}{(n-r)(n-r+1)}$ are positive, which implies $k^2-8$ must be positive.