Key Concept: Binomial Theorem and Coefficient Comparison

This problem leverages the Binomial Theorem to expand polynomial expressions and then requires careful algebraic manipulation to compare coefficients of specific powers of $y$. The general binomial expansion for $(1+x)^k$ up to the $x^2$ term is crucial:

$(1+x)^k = \binom{k}{0}x^0 + \binom{k}{1}x^1 + \binom{k}{2}x^2 + \dots$ $(1+x)^k = 1 + kx + \frac{k(k-1)}{2!}x^2 + \dots$ Where $\binom{k}{r} = \frac{k!}{r!(k-r)!}$ represents the binomial coefficient. This formula is applicable for any real $k$, and for natural numbers $k$, the series is finite.

Step 1: Expanding the Individual Binomial Terms

We need to expand both $(1-y)^m$ and $(1+y)^n$ up to the $y^2$ term, as the problem provides information only about the coefficients of $y^1$ ($a_1$) and $y^2$ ($a_2$) in the product. Expanding beyond $y^2$ would be unnecessary for determining $a_1$ and $a_2$.

1. For $(1-y)^m$: We substitute $x = -y$ and $k = m$ into the binomial expansion formula. The negative sign for $y$ means that terms with odd powers of $y$ will have a negative sign, and terms with even powers of $y$ will have a positive sign. $(1-y)^m = 1 + m(-y) + \frac{m(m-1)}{2!}(-y)^2 + O(y^3)$ $(1-y)^m = 1 - my + \frac{m(m-1)}{2}y^2 + O(y^3) \quad \text{(Equation A)}$ Explanation: The binomial coefficients are applied to $y$, and the sign is determined by the power of $(-y)$. For instance, $(-y)^1 = -y$ and $(-y)^2 = y^2$.

2. For $(1+y)^n$: We substitute $x = y$ and $k = n$ into the binomial expansion formula. Since $y$ is positive, all terms in this expansion will be positive. $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + O(y^3)$ $(1+y)^n = 1 + ny + \frac{n(n-1)}{2}y^2 + O(y^3) \quad \text{(Equation B)}$ Explanation: The standard binomial expansion is applied directly.

Step 2: Multiplying Expansions and Extracting Coefficients

Now, we multiply the two expanded series (Equation A and Equation B) and equate the result to the given expression $1 + {a_1}y + {a_2}{y^2} + \dots$. We only need to consider terms that contribute to $y^1$ and $y^2$.

$\left( 1 - my + \frac{m(m-1)}{2}y^2 + \dots \right) \left( 1 + ny + \frac{n(n-1)}{2}y^2 + \dots \right) = 1 + {a_1}y + {a_2}{y^2} + \dots$

1. Coefficient of $y$ ($a_1$): To find the coefficient of $y$, we identify all pairs of terms from the two series whose product results in a $y$ term.

   • (Constant from first series) $\times$ ($y$ term from second series): $1 \cdot (ny) = ny$
   • ($y$ term from first series) $\times$ (Constant from second series): $(-my) \cdot 1 = -my$ Summing these, the coefficient of $y$ is $a_1 = n - m$.

2. Coefficient of $y^2$ ($a_2$): To find the coefficient of $y^2$, we identify all pairs of terms from the two series whose product results in a $y^2$ term.

   • (Constant from first series) $\times$ ($y^2$ term from second series): $1 \cdot \left( \frac{n(n-1)}{2}y^2 \right)$
   • ($y$ term from first series) $\times$ ($y$ term from second series): $(-my) \cdot (ny) = -mny^2$
   • ($y^2$ term from first series) $\times$ (Constant from second series): $\left( \frac{m(m-1)}{2}y^2 \right) \cdot 1$ Summing these, the coefficient of $y^2$ is $a_2 = \frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2}$.

Step 3: Setting Up and Solving the System of Equations

We are given that $a_1 = 10$ and $a_2 = 10$. This allows us to form a system of two equations with two variables, $m$ and $n$:

1. From $a_1$: $n - m = 10 \quad \text{(Equation 1)}$ Explanation: This equation is straightforward and directly relates $n$ and $m$. It's a good starting point for substitution.

2. From $a_2$: $\frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2} = 10 \quad \text{(Equation 2)}$

Solving for $n$ in terms of $m$ (from Equation 1): From Equation 1, we can easily express $n$ as: $n = m + 10$ Explanation: This step isolates $n$ to enable its substitution into Equation 2, simplifying Equation 2 to contain only one variable, $m$.

Substituting $n$ into Equation 2: Substitute $n = m + 10$ into Equation 2: $\frac{(m+10)((m+10)-1)}{2} - m(m+10) + \frac{m(m-1)}{2} = 10$ $\frac{(m+10)(m+9)}{2} - m(m+10) + \frac{m(m-1)}{2} = 10$

Clearing Denominators: To simplify the equation and eliminate fractions, multiply the entire equation by 2: $2 \times \left( \frac{(m+10)(m+9)}{2} - m(m+10) + \frac{m(m-1)}{2} \right) = 2 \times 10$ $(m+10)(m+9) - 2m(m+10) + m(m-1) = 20$ Explanation: Multiplying by the common denominator (2 in this case) is a standard algebraic technique to simplify equations with fractions, making subsequent expansions less error-prone.

Expanding and Simplifying the Equation: Now, we expand each product term:

• $(m+10)(m+9) = m^2 + 9m + 10m + 90 = m^2 + 19m + 90$
• $-2m(m+10) = -2m^2 - 20m$
• $m(m-1) = m^2 - m$

Substitute these expanded terms back into the equation: $(m^2 + 19m + 90) + (-2m^2 - 20m) + (m^2 - m) = 20$

Combine like terms (terms with $m^2$, terms with $m$, and constant terms): $(m^2 - 2m^2 + m^2) + (19m - 20m - m) + 90 = 20$ $(1 - 2 + 1)m^2 + (19 - 20 - 1)m + 90 = 20$ $0m^2 + (-2)m + 90 = 20$ $-2m + 90 = 20$ Explanation: Notice that the $m^2$ terms perfectly cancel out ($m^2 - 2m^2 + m^2 = 0m^2$). This simplifies the equation from a potential quadratic to a linear equation in $m$, which is much easier to solve.

Solving for $m$: Isolate $m$ by performing standard algebraic operations: $-2m = 20 - 90$ $-2m = -70$ $m = \frac{-70}{-2}$ $m = 35$ Explanation: Subtract 90 from both sides, then divide by -2 to find the value of $m$.

Solving for $n$: Now that we have the value of $m$, use Equation 1 ($n = m + 10$) to find $n$: $n = 35 + 10$ $n = 45$ Explanation: Substitute the calculated value of $m$ back into the simpler Equation 1 to find $n$.

Step 4: Calculating the Value of $(m+n)$

The problem asks for the value of $(m+n)$: $m + n = 35 + 45 = 80$

Tips and Common Mistakes to Avoid

• Sign Errors: When dealing with $(1-y)^m$, pay close attention to the alternating signs for the terms. A common mistake is to treat it as $(1+y)^m$.
• Thorough Multiplication: When multiplying two polynomial expansions, ensure that every term from the first expansion is multiplied by every term from the second expansion up to the required power of $y$. A systematic approach (e.g., creating a small table or multiplying term-by-term) can prevent missing terms.
• Algebraic Precision: Algebraic manipulations, especially when expanding products and combining like terms, are a frequent source of errors. Double-check each step.
• Focus on Relevant Terms: Only expand and multiply terms that will contribute to the coefficients you need ($a_1$ and $a_2$ in this case). This saves time and reduces complexity.
• Alternative for Coefficients (Advanced): For higher powers or complex scenarios, recalling that $a_k = \frac{f^{(k)}(0)}{k!}$ where $f(y) = (1-y)^m(1+y)^n$ can be a useful verification tool, involving derivatives.

Summary and Key Takeaway

By meticulously applying the Binomial Theorem to expand $(1-y)^m$ and $(1+y)^n$ up to the $y^2$ term, then carefully multiplying these series, we were able to equate the coefficients of $y$ and $y^2$ to the given values of $a_1=10$ and $a_2=10$. This led to a system of two linear equations. Solving this system yielded $m=35$ and $n=45$. Therefore, the value of $(m+n)$ is $35+45=80$. This problem underscores the importance of precise binomial expansion and careful, step-by-step algebraic simplification.

Note on Discrepancy: Based on the detailed calculations following standard mathematical principles, the derived value for $(m+n)$ is $80$. If the intended correct option for the problem was (A) 88, there might be a typographical error in the provided problem statement or options, as our derived answer is consistently 80.