If the coefficient of `x^(2)`+ coefficient of x in the expanssion of `(1+x)^(m)(1-x)^(n),(mnen)` is equal to – m, then the value of n – m is equal to:

Answer

To solve the problem, we need to find the value of \( n - m \) given that the coefficient of \( x^2 \) plus the coefficient of \( x \) in the expansion of \( (1+x)^m(1-x)^n \) is equal to \( -m \).

### Step 1: Find the coefficient of \( x \) in the expansion
The coefficient of \( x \) in \( (1+x)^m \) is \( \binom{m}{1} = m \) and in \( (1-x)^n \) is \( -\binom{n}{1} = -n \). Therefore, the coefficient of \( x \) in the product \( (1+x)^m(1-x)^n \) is:
\[ \text{Coefficient of } x = m - n \]

### Step 2: Find the coefficient of \( x^2 \) in the expansion
The coefficient of \( x^2 \) in \( (1+x)^m \) is \( \binom{m}{2} = \frac{m(m-1)}{2} \) and in \( (1-x)^n \) is \( \binom{n}{2} = \frac{n(n-1)}{2} \). The coefficient of \( x^2 \) in the product \( (1+x)^m(1-x)^n \) is:
\[ \text{Coefficient of } x^2 = \frac{m(m-1)}{2} + \frac{n(n-1)}{2} - mn \]
This simplifies to:
\[ \text{Coefficient of } x^2 = \frac{m(m-1) + n(n-1) - 2mn}{2} = \frac{m^2 - m + n^2 - n - 2mn}{2} \]

### Step 3: Set up the equation
According to the problem, the sum of the coefficients of \( x^2 \) and \( x \) is equal to \( -m \):
\[ \frac{m^2 - m + n^2 - n - 2mn}{2} + (m - n) = -m \]
Multiplying through by 2 to eliminate the fraction:
\[ m^2 - m + n^2 - n - 2mn + 2m - 2n = -2m \]
This simplifies to:
\[ m^2 + n^2 - 2mn + m - 3n = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ m^2 - 2mn + n^2 + m - 3n = 0 \]
This can be factored as:
\[ (m - n)^2 + m - 3n = 0 \]

### Step 5: Solve for \( n - m \)
Let \( k = n - m \). Then, substituting \( n = m + k \):
\[ (m - (m + k))^2 + m - 3(m + k) = 0 \]
This simplifies to:
\[ (-k)^2 + m - 3m - 3k = 0 \]
\[ k^2 - 2m - 3k = 0 \]
Factoring gives:
\[ k(k - 3) = 2m \]
This means \( k = 3 \) or \( k = 0 \). Since \( m \neq n \), we take \( k = 3 \).

Thus, we find:
\[ n - m = 3 \]

### Final Answer
The value of \( n - m \) is \( 3 \).

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