IF `""^(n-1)C_r=(k^2-3)"^n C_(r+1)` then `k in `

Options

A. `(-sqrt3,sqrt3)`
B. `(sqrt3,2)`
C. `(0,sqrt3)`
D. `(sqrt3,2)`

Answer

To solve the equation \( \binom{n-1}{r} = (k^2 - 3) \binom{n}{r+1} \), we will follow these steps:

### Step 1: Write the formulas for combinations
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Using this, we can express \( \binom{n-1}{r} \) and \( \binom{n}{r+1} \).

### Step 2: Express \( \binom{n-1}{r} \)
Using the formula:
\[ \binom{n-1}{r} = \frac{(n-1)!}{r!(n-1-r)!} \]

### Step 3: Express \( \binom{n}{r+1} \)
Using the formula:
\[ \binom{n}{r+1} = \frac{n!}{(r+1)!(n-(r+1))!} = \frac{n!}{(r+1)!(n-r-1)!} \]

### Step 4: Rewrite \( n! \)
We can rewrite \( n! \) as:
\[ n! = n \cdot (n-1)! \]
Substituting this into the expression for \( \binom{n}{r+1} \):
\[ \binom{n}{r+1} = \frac{n \cdot (n-1)!}{(r+1)!(n-r-1)!} \]

### Step 5: Substitute into the original equation
Now substituting both expressions into the original equation:
\[ \frac{(n-1)!}{r!(n-1-r)!} = (k^2 - 3) \cdot \frac{n \cdot (n-1)!}{(r+1)!(n-r-1)!} \]

### Step 6: Cancel \( (n-1)! \)
We can cancel \( (n-1)! \) from both sides:
\[ \frac{1}{r!(n-1-r)!} = (k^2 - 3) \cdot \frac{n}{(r+1)!(n-r-1)!} \]

### Step 7: Simplify the equation
Rearranging gives:
\[ \frac{(r+1)!}{r!(n-1-r)!} = (k^2 - 3) \cdot \frac{n}{(n-r-1)!} \]
This simplifies to:
\[ \frac{r+1}{n-r} = (k^2 - 3) \cdot \frac{n}{(n-r-1)!} \]

### Step 8: Isolate \( k^2 \)
From the above equation, we can isolate \( k^2 \):
\[ k^2 - 3 = \frac{(r+1)(n-r-1)!}{n(n-r)} \]
Thus,
\[ k^2 = \frac{(r+1)(n-r-1)!}{n(n-r)} + 3 \]

### Step 9: Determine the range for \( k \)
To ensure \( \binom{n-1}{r} \) is valid, we need \( r \leq n-1 \) and \( r \geq 0 \). This implies:
\[ 1 \leq r + 1 \leq n \]
Dividing by \( n \):
\[ \frac{1}{n} \leq \frac{r+1}{n} \leq 1 \]
Adding 3 gives:
\[ \frac{1}{n} + 3 \leq k^2 \leq 4 \]
Taking square roots:
\[ \sqrt{\frac{1}{n} + 3} \leq k \leq 2 \]

### Final Result
As \( n \) approaches infinity, \( \frac{1}{n} \) approaches 0, leading to:
\[ \sqrt{3} \leq k \leq 2 \]
Thus, the value of \( k \) belongs to the interval:
\[ k \in [\sqrt{3}, 2] \]