If `r^[th]` and `(r+1)^[th]` term in the expansion of `(p+q)^n` are equal, then `[(n+1)q]/[r(p+q)]` is (a) `1/2` (b) `1/4` (c) 1 (d) 0

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `1`
D. `0`

Explanation

To solve the problem step by step, we need to analyze the given information and apply the Binomial Theorem.

### Step 1: Understand the Terms
We know that the \( r^{th} \) term and the \( (r+1)^{th} \) term in the expansion of \( (p+q)^n \) are equal. According to the Binomial Theorem, the \( r^{th} \) term \( T_r \) is given by:
\[ T_r = \binom{n}{r} p^{n-r} q^r \]
And the \( (r+1)^{th} \) term \( T_{r+1} \) is:
\[ T_{r+1} = \binom{n}{r+1} p^{n-(r+1)} q^{r+1} \]

### Step 2: Set the Terms Equal
Since \( T_r = T_{r+1} \), we can set up the equation:
\[ \binom{n}{r} p^{n-r} q^r = \binom{n}{r+1} p^{n-(r+1)} q^{r+1} \]

### Step 3: Substitute Binomial Coefficients
Recall that:
\[ \binom{n}{r+1} = \frac{n-r}{r+1} \binom{n}{r} \]
Substituting this into the equation gives:
\[ \binom{n}{r} p^{n-r} q^r = \frac{n-r}{r+1} \binom{n}{r} p^{n-r-1} q^{r+1} \]

### Step 4: Cancel Common Terms
Assuming \( \binom{n}{r} \neq 0 \), we can divide both sides by \( \binom{n}{r} \):
\[ p^{n-r} q^r = \frac{n-r}{r+1} p^{n-r-1} q^{r+1} \]
Now, we can simplify:
\[ p q^r = \frac{n-r}{r+1} q^{r+1} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ p = \frac{n-r}{r+1} q \]

### Step 6: Cross-Multiply
Cross-multiplying gives:
\[ p(r+1) = (n-r)q \]

### Step 7: Isolate \( n+1 \)
Rearranging this equation leads to:
\[ pr + p = nq - rq \]
This can be rewritten as:
\[ pr + rq + p = nq \]
Factoring out \( q \) gives:
\[ (n+1)q = r(p+q) \]

### Step 8: Solve for the Required Expression
Now, we need to find:
\[ \frac{(n+1)q}{r(p+q)} \]
From our previous step, we have:
\[ \frac{(n+1)q}{r(p+q)} = 1 \]

### Final Answer
Thus, the value of \( \frac{(n+1)q}{r(p+q)} \) is:
\[ \boxed{1} \]