Understanding the Problem and Key Concept

The problem asks us to find the ratio $\frac{a_2}{a_0}$ from the polynomial expansion of $(x + 10)^{50} + (x - 10)^{50}$. The expanded form is given as $a_0 + a_1 x + a_2 x^2 + \dots + a_{50} x^{50}$. This means we need to find the constant term ($a_0$) and the coefficient of $x^2$ ($a_2$) in the sum of these two binomial expansions.

The fundamental tool we will use here is the Binomial Theorem, which states that for any non-negative integer $n$: $(A+B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k = \binom{n}{0}A^n + \binom{n}{1}A^{n-1}B + \binom{n}{2}A^{n-2}B^2 + \dots + \binom{n}{n}B^n$ And for $(A-B)^n$: $(A-B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} (-B)^k = \binom{n}{0}A^n - \binom{n}{1}A^{n-1}B + \binom{n}{2}A^{n-2}B^2 - \dots + (-1)^n \binom{n}{n}B^n$ A key observation for this problem is what happens when we add these two expansions: $(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0}A^n + \binom{n}{2}A^{n-2}B^2 + \binom{n}{4}A^{n-4}B^4 + \dots \right]$ The terms with odd powers of $B$ cancel out due to the alternating signs in the $(A-B)^n$ expansion. This simplification is crucial for efficiently finding specific coefficients.

In our problem, the expression is $(x + 10)^{50} + (x - 10)^{50}$. To make it easier to identify the coefficients $a_0$ (constant term) and $a_2$ (coefficient of $x^2$), it's often helpful to express the binomials in the form $(C + Var)^n$, where $C$ is a constant and $Var$ is the variable term. So we can rewrite the terms as $(10 + x)^{50}$ and $(10 - x)^{50}$. Here, $A = 10$, $B = x$, and $n = 50$.

Step-by-Step Derivation

1. Expand $(10+x)^{50}$ using the Binomial Theorem: We need terms up to $x^2$. $(10+x)^{50} = \binom{50}{0}10^{50}x^0 + \binom{50}{1}10^{49}x^1 + \binom{50}{2}10^{48}x^2 + \dots$

   • Explanation: We apply the binomial theorem with $A=10$, $B=x$, and $n=50$. The general term is $\binom{n}{k}A^{n-k}B^k$. We are interested in terms where $x$ has power 0 (for $a_0$) and power 2 (for $a_2$).

2. Expand $(10-x)^{50}$ using the Binomial Theorem: Similarly, $(10-x)^{50} = \binom{50}{0}10^{50}(-x)^0 + \binom{50}{1}10^{49}(-x)^1 + \binom{50}{2}10^{48}(-x)^2 + \dots$ $(10-x)^{50} = \binom{50}{0}10^{50} - \binom{50}{1}10^{49}x + \binom{50}{2}10^{48}x^2 - \dots$

   • Explanation: Here $A=10$, $B=-x$, and $n=50$. The odd powers of $-x$ will result in negative terms, while even powers will remain positive.

3. Sum the two expansions: Now, we add the expansions of $(10+x)^{50}$ and $(10-x)^{50}$: $(10+x)^{50} + (10-x)^{50} = \left( \binom{50}{0}10^{50} + \binom{50}{1}10^{49}x + \binom{50}{2}10^{48}x^2 + \dots \right) + \left( \binom{50}{0}10^{50} - \binom{50}{1}10^{49}x + \binom{50}{2}10^{48}x^2 - \dots \right)$ Combining like terms: $(10+x)^{50} + (10-x)^{50} = 2 \left[ \binom{50}{0}10^{50} + \binom{50}{2}10^{48}x^2 + \binom{50}{4}10^{46}x^4 + \dots \right]$

   • Explanation: As discussed earlier, the terms with odd powers of $x$ (like $x^1, x^3, \dots$) cancel each other out, while terms with even powers of $x$ (like $x^0, x^2, x^4, \dots$) are doubled.

4. Identify $a_0$ and $a_2$: The given polynomial is $a_0 + a_1 x + a_2 x^2 + \dots + a_{50} x^{50}$. By comparing this with our sum:

   • The constant term ($x^0$ term) is $a_0$: $a_0 = 2 \times \binom{50}{0}10^{50}$ Since $\binom{50}{0} = 1$: $a_0 = 2 \times 1 \times 10^{50} = 2 \times 10^{50}$
   • The coefficient of $x^2$ is $a_2$: $a_2 = 2 \times \binom{50}{2}10^{48}$
   • Explanation: We are directly extracting the coefficients for $x^0$ and $x^2$ from the combined expansion, matching them to the definition of $a_0$ and $a_2$.

5. Calculate $\binom{50}{2}$: The combination formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. $\binom{50}{2} = \frac{50 \times 49}{2 \times 1} = 25 \times 49$ $25 \times 49 = 1225$

   • Explanation: This is a standard calculation for binomial coefficients. It's often quicker to calculate $\binom{n}{2}$ as $\frac{n(n-1)}{2}$.

6. Compute the ratio $\frac{a_2}{a_0}$: Now substitute the values of $a_0$ and $a_2$: $\frac{a_2}{a_0} = \frac{2 \times \binom{50}{2}10^{48}}{2 \times 10^{50}}$ Cancel out the common factor of 2: $\frac{a_2}{a_0} = \frac{\binom{50}{2}10^{48}}{10^{50}}$ Using the rules of exponents ($x^m / x^n = x^{m-n}$): $\frac{a_2}{a_0} = \frac{\binom{50}{2}}{10^{50-48}}$ $\frac{a_2}{a_0} = \frac{\binom{50}{2}}{10^2}$ Substitute the value of $\binom{50}{2}$ and $10^2$: $\frac{a_2}{a_0} = \frac{1225}{100}$ $\frac{a_2}{a_0} = 12.25$

   • Explanation: We substitute the derived expressions for $a_2$ and $a_0$ into the required ratio and simplify using basic algebraic and exponent rules.

Important Tips and Common Mistakes

• Order of terms in binomial: Be careful when identifying $A$ and $B$. If the question had been $(10+x)^{50} + (x-10)^{50}$, the second term $(x-10)^{50}$ would be $(-(10-x))^{50}$, which is $(10-x)^{50}$ since $50$ is an even power. However, directly expanding $(x-10)^{50}$ with $A=x, B=10$ is also valid, but then you'd be looking for the coefficient of $x^2$ in the first term's expansion and $x^2$ in the second term's expansion, and then adding them. Rewriting to have the variable as the second term (e.g., $(10+x)$) often simplifies coefficient identification for polynomials in $x$.
• Sign errors: When dealing with $(A-B)^n$, remember that $(-B)^k$ is negative for odd $k$ and positive for even $k$.
• Binomial coefficient calculation: Double-check your calculations for $\binom{n}{k}$, especially for larger $n$.
• Exponent rules: Be precise with exponent operations, particularly when dividing terms with the same base.
• Understanding the question: Ensure you are calculating the correct coefficients ($a_0$ for $x^0$, $a_2$ for $x^2$) and not mixing them up.

Summary/Key Takeaway

This problem effectively tests your understanding of the Binomial Theorem and your ability to apply it to find specific coefficients in a polynomial expansion. The key insight is recognizing the symmetry in $(A+B)^n + (A-B)^n$, which allows for significant simplification by canceling out odd-powered terms. This reduces the computational effort and minimizes chances of error. The ratio of coefficients can then be found by carefully extracting and simplifying the relevant terms.