If `(""^n P_(r-1))/a=(""^n P_r)/b=("^n P_(r+1))/c` then

Options

A. `sum 1/a=1`
B. `abc=1`
C. `b^2=a(b+c)`
D. `a^2=c(a+b)`

Answer

To solve the equation given by the condition \((\binom{n}{r-1}/a) = (\binom{n}{r}/b) = (\binom{n}{r+1}/c)\), we will derive a relationship between \(A\), \(B\), and \(C\).

### Step 1: Set up the equations
Given:
\[ \frac{\binom{n}{r-1}}{a} = \frac{\binom{n}{r}}{b} = \frac{\binom{n}{r+1}}{c} = k \text{ (some constant)} \]

From this, we can write:
\[ \binom{n}{r-1} = ak, \quad \binom{n}{r} = bk, \quad \binom{n}{r+1} = ck \]

### Step 2: Use the formula for combinations
Recall the formula for combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Using this, we can express \(\binom{n}{r-1}\) and \(\binom{n}{r+1}\):
\[ \binom{n}{r-1} = \frac{n!}{(r-1)!(n-(r-1))!} = \frac{n!}{(r-1)!(n-r+1)!} \]
\[ \binom{n}{r+1} = \frac{n!}{(r+1)!(n-(r+1))!} = \frac{n!}{(r+1)!(n-r-1)!} \]

### Step 3: Relate the combinations
We can relate \(\binom{n}{r}\), \(\binom{n}{r-1}\), and \(\binom{n}{r+1}\) using the following relationships:
\[ \binom{n}{r} = \frac{n-r+1}{r} \cdot \binom{n}{r-1} \]
\[ \binom{n}{r+1} = \frac{r+1}{n-r} \cdot \binom{n}{r} \]

### Step 4: Substitute into the equations
From the first equation:
\[ \frac{ak}{b} = \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{r}{n-r+1} \]
Cross-multiplying gives:
\[ b = a \cdot k \cdot \frac{n-r+1}{r} \]

From the second equation:
\[ \frac{bk}{c} = \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} \]
Cross-multiplying gives:
\[ c = b \cdot k \cdot \frac{r+1}{n-r} \]

### Step 5: Substitute \(b\) into the equation for \(c\)
Substituting the expression for \(b\) into the equation for \(c\):
\[ c = \left(a \cdot k \cdot \frac{n-r+1}{r}\right) \cdot k \cdot \frac{r+1}{n-r} \]
This simplifies to:
\[ c = a \cdot k^2 \cdot \frac{(n-r+1)(r+1)}{r(n-r)} \]

### Step 6: Establish the final relationship
Now we have expressions for \(b\) and \(c\) in terms of \(a\) and \(k\). We can express the relationship:
\[ b^2 = a(b + c) \]

### Conclusion
Thus, the final condition derived from the given expression is:
\[ b^2 = a(b + c) \]