Rewritten Solution

1. Key Concept: Binomial Theorem and Series Multiplication

This problem utilizes the Binomial Theorem for expanding expressions of the form $(a+b)^n$ and then finding coefficients of specific terms in the product of two such expansions.

The Binomial Theorem states that for any natural number $n$, $(1+x)^n = \sum_{k=0}^{n} {}^n{C_k} x^k = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}x^2 + {}^n{C_3}x^3 + \ldots + {}^n{C_n}x^n$ where ${}^n{C_k} = \frac{n!}{k!(n-k)!}$. Similarly, for $(1-x)^m$, we have: $(1-x)^m = {}^m{C_0} - {}^m{C_1}x + {}^m{C_2}x^2 - {}^m{C_3}x^3 + \ldots + (-1)^m{}^m{C_m}x^m$ Notice the alternating signs for the expansion of $(1-x)^m$. This is a common point of error.

We are given the product: ${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + \ldots$ To find ${a_1}$ (coefficient of $y$) and ${a_2}$ (coefficient of ${y^2}$), we need to expand each factor up to the ${y^2}$ term and then multiply the series.

The expansions up to the ${y^2}$ term are: $(1-y)^m = {}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + O(y^3) = 1 - my + \frac{m(m-1)}{2}{y^2} + O(y^3)$ $(1+y)^n = {}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + O(y^3) = 1 + ny + \frac{n(n-1)}{2}{y^2} + O(y^3)$ Here, $O(y^3)$ represents terms of order ${y^3}$ and higher, which we do not need for finding ${a_1}$ and ${a_2}$.

2. Calculating the Coefficient of $y$ (${a_1}$)

To find the coefficient of $y$ in the product ${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}$, we multiply the expanded series: $\left( {1 - my + \frac{m(m-1)}{2}{y^2} + \ldots} \right) \left( {1 + ny + \frac{n(n-1)}{2}{y^2} + \ldots} \right)$ The terms that will produce $y$ are:

• $(1) \cdot (ny)$ from multiplying the constant term of the first series by the $y$ term of the second series. This gives $ny$.
• $(-my) \cdot (1)$ from multiplying the $y$ term of the first series by the constant term of the second series. This gives $-my$.

Combining these, the coefficient of $y$ (${a_1}$) is: ${a_1} = n - m$ We are given that ${a_1} = 10$. Therefore, we have our first equation: $n - m = 10 \quad \ldots(1)$

3. Calculating the Coefficient of ${y^2}$ (${a_2}$)

To find the coefficient of ${y^2}$ in the product, we identify all combinations of terms that multiply to form ${y^2}$:

• $(1) \cdot \left( \frac{n(n-1)}{2}{y^2} \right)$ from constant term $\times$ ${y^2}$ term. This gives $\frac{n(n-1)}{2}{y^2}$.
• $(-my) \cdot (ny)$ from $y$ term $\times$ $y$ term. This gives $-mn{y^2}$.
• $\left( \frac{m(m-1)}{2}{y^2} \right) \cdot (1)$ from ${y^2}$ term $\times$ constant term. This gives $\frac{m(m-1)}{2}{y^2}$.

Combining these, the coefficient of ${y^2}$ (${a_2}$) is: ${a_2} = \frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2}$ We are given that ${a_2} = 10$. So, our second equation is: $\frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2} = 10$ To simplify, multiply the entire equation by 2: $n(n-1) - 2mn + m(m-1) = 20 \quad \ldots(2)$

Common Mistake Alert: A frequent error is mismanaging the signs when terms from $(1-y)^m$ are involved, particularly the $-my$ term. Ensure each combination carries its correct sign.

4. Solving the System of Equations

We have a system of two linear equations in two variables, $m$ and $n$:

1. $n - m = 10$
2. $n(n-1) - 2mn + m(m-1) = 20$

From equation (1), we can express $n$ in terms of $m$: $n = m + 10$ Now, substitute this expression for $n$ into equation (2): $(m+10)( (m+10)-1 ) - 2m(m+10) + m(m-1) = 20$ $(m+10)(m+9) - 2m(m+10) + m(m-1) = 20$ Now, expand and simplify the terms: $(m^2 + 9m + 10m + 90) - (2m^2 + 20m) + (m^2 - m) = 20$ $(m^2 + 19m + 90) - 2m^2 - 20m + m^2 - m = 20$ Collect like terms ($m^2$, $m$, and constant terms): $(m^2 - 2m^2 + m^2) + (19m - 20m - m) + (90) = 20$ $(0)m^2 + (-2m) + 90 = 20$ $-2m + 90 = 20$ Subtract 90 from both sides: $-2m = 20 - 90$ $-2m = -70$ Divide by -2: $m = \frac{-70}{-2}$ $m = 35$

Now that we have the value of $m$, substitute it back into equation (1) to find $n$: $n = m + 10$ $n = 35 + 10$ $n = 45$

Thus, the values are $m = 35$ and $n = 45$.

5. Verification

Let's quickly verify our solution using the calculated values for $m$ and $n$: Equation (1): $n - m = 45 - 35 = 10$. This matches the given ${a_1} = 10$. Equation (2): ${a_2} = \frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2}$ ${a_2} = \frac{45(44)}{2} - (35)(45) + \frac{35(34)}{2}$ ${a_2} = 45 \cdot 22 - 1575 + 35 \cdot 17$ ${a_2} = 990 - 1575 + 595$ ${a_2} = 1585 - 1575$ ${a_2} = 10$ This also matches the given ${a_2} = 10$. The values are consistent.

Therefore, the pair $(m, n)$ is $(35, 45)$.

6. Summary and Key Takeaway

This problem demonstrates a classic application of the Binomial Theorem combined with algebraic manipulation. The key steps involve:

1. Expanding Binomials: Correctly expanding each binomial factor up to the required power (here, ${y^2}$), paying close attention to signs, especially for $(1-y)^m$.
2. Coefficient Extraction: Systematically identifying and summing the terms that contribute to the desired power of $y$ in the product of the series.
3. Solving Simultaneous Equations: Setting up and solving the system of equations derived from the given coefficients.

Always double-check your algebraic simplifications to avoid calculation errors, which are common in such multi-step problems.