For natural numbers m, n if `(1-y)^(m)(1+y)^(n) = 1+a_(1)y+a_(2)y^(2) + "……."` and `a_(1) = a_(2) = 10`, then `(m,n)` is :

Options

A. (35,20)
B. (45,35)
C. (35,45)
D. (20,45)

Answer

To solve the problem, we need to analyze the equation given and extract the coefficients of \(y\) and \(y^2\) from the expression \((1-y)^m(1+y)^n\).

### Step-by-step Solution:

1. **Expand the Expression:**
We start with the expression:
\[ (1-y)^m(1+y)^n \]
Using the binomial theorem, we can expand both parts:
\[ (1-y)^m = \sum_{k=0}^{m} \binom{m}{k} (-y)^k = \sum_{k=0}^{m} \binom{m}{k} (-1)^k y^k \]
\[ (1+y)^n = \sum_{j=0}^{n} \binom{n}{j} y^j \]

2. **Combine the Expansions:**
The product of these two expansions gives:
\[ (1-y)^m(1+y)^n = \sum_{k=0}^{m} \sum_{j=0}^{n} \binom{m}{k} \binom{n}{j} (-1)^k y^{k+j} \]

3. **Identify Coefficients:**
We need to find the coefficients of \(y\) and \(y^2\):
- The coefficient of \(y\) is obtained when \(k+j=1\):
\[ a_1 = \binom{m}{1} \binom{n}{0} (-1)^1 + \binom{m}{0} \binom{n}{1} = -m + n \]
- The coefficient of \(y^2\) is obtained when \(k+j=2\):
\[ a_2 = \binom{m}{2} \binom{n}{0} (-1)^2 + \binom{m}{1} \binom{n}{1} (-1)^1 + \binom{m}{0} \binom{n}{2} \]
Simplifying this gives:
\[ a_2 = \frac{m(m-1)}{2} - mn + \frac{n(n-1)}{2} \]

4. **Set Up Equations:**
From the problem statement, we know:
\[ a_1 = n - m = 10 \quad \text{(Equation 1)} \]
\[ a_2 = \frac{m(m-1)}{2} - mn + \frac{n(n-1)}{2} = 10 \quad \text{(Equation 2)} \]

5. **Substitute \(n\) in Terms of \(m\):**
From Equation 1, we can express \(n\) as:
\[ n = m + 10 \]

6. **Substitute into Equation 2:**
Substitute \(n\) into Equation 2:
\[ \frac{m(m-1)}{2} - m(m+10) + \frac{(m+10)(m+9)}{2} = 10 \]
Simplifying this gives:
\[ \frac{m(m-1) - 2m(m+10) + (m^2 + 19m + 90)}{2} = 10 \]
Multiply through by 2 to eliminate the fraction:
\[ m(m-1) - 2m(m+10) + m^2 + 19m + 90 = 20 \]
Combine like terms:
\[ -m^2 - 21m + 90 = 20 \]
Rearranging gives:
\[ m^2 + 21m + 70 = 0 \]

7. **Solve the Quadratic Equation:**
Using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-21 \pm \sqrt{441 - 280}}{2} = \frac{-21 \pm \sqrt{161}}{2} \]
Since \(m\) must be a natural number, we can check integer values around the roots.

8. **Find \(m\) and \(n\):**
Testing \(m = 35\) gives:
\[ n = 35 + 10 = 45 \]

Thus, the solution is:
\[ (m, n) = (35, 45) \]