In the expansion of `(a+b)^(n)`, if two consecutive terms are equal, then which of the following is/are always integer ?
(a)`((n+1)b)/(a+b)` (b)`((n+1)a)/(a+b)` (c)`(na)/(a-b)` (d)`(na)/(a+b)`

Options

A. `((n+1)b)/(a+b)`
B. `((n+1)a)/(a+b)`
C. `(na)/(a-b)`
D. `(na)/(a+b)`

Explanation

To solve the problem, we need to analyze the condition when two consecutive terms in the binomial expansion of \((a + b)^n\) are equal. Let's go through the steps systematically.

### Step 1: Identify the Terms
In the binomial expansion of \((a + b)^n\), the \(r\)-th term \(T_r\) is given by:
\[ T_r = \binom{n}{r-1} a^{n - (r - 1)} b^{(r - 1)} = \binom{n}{r-1} a^{n - r + 1} b^{r - 1} \]

### Step 2: Set Up the Equation for Consecutive Terms
If two consecutive terms are equal, we have:
\[ T_r = T_{r+1} \]
This translates to:
\[ \binom{n}{r-1} a^{n - r + 1} b^{r - 1} = \binom{n}{r} a^{n - r} b^r \]

### Step 3: Simplify the Equation
Using the property of binomial coefficients, we can rewrite the equation:
\[ \binom{n}{r-1} = \frac{n!}{(r-1)!(n - (r - 1))!} \quad \text{and} \quad \binom{n}{r} = \frac{n!}{r!(n - r)!} \]
Thus, we can express the equality as:
\[ \frac{n!}{(r-1)!(n - r + 1)!} a^{n - r + 1} b^{r - 1} = \frac{n!}{r!(n - r)!} a^{n - r} b^r \]

### Step 4: Cancel Common Terms
Canceling \(n!\) from both sides, we have:
\[ \frac{1}{(r-1)!(n - r + 1)!} a^{n - r + 1} b^{r - 1} = \frac{1}{r!(n - r)!} a^{n - r} b^r \]
This simplifies to:
\[ \frac{a^{n - r + 1} b^{r - 1}}{(r-1)!(n - r + 1)!} = \frac{a^{n - r} b^r}{r!(n - r)!} \]

### Step 5: Cross-Multiply
Cross-multiplying gives:
\[ a^{n - r + 1} b^{r - 1} \cdot r! (n - r)! = a^{n - r} b^r \cdot (r - 1)! (n - r + 1)! \]

### Step 6: Rearranging the Equation
This leads us to:
\[ a \cdot r = b \cdot (n - r + 1) \]
From this, we can express \(r\):
\[ r = \frac{b(n + 1)}{a + b} \]

### Step 7: Check Integer Conditions
Since \(r\) must be an integer, the right-hand side must also be an integer. This implies that \(b(n + 1)\) must be divisible by \(a + b\).

### Step 8: Analyze the Options
Now, we check the given options to see which expressions are always integers:

(a) \(\frac{(n + 1)b}{a + b}\)
(b) \(\frac{(n + 1)a}{a + b}\)
(c) \(\frac{na}{a - b}\)
(d) \(\frac{na}{a + b}\)

Since \(r\) is an integer, we can conclude that:
- Option (a) is an integer because it is derived from the integer \(r\).
- Option (b) is not guaranteed to be an integer.
- Option (c) is not guaranteed to be an integer.
- Option (d) is also an integer since it is derived from the same conditions.

### Final Conclusion
The only option that is always an integer is:
- (a) \(\frac{(n + 1)b}{a + b}\)