Let `(x+10)^(50)+(x-10)^(50)=a_(0)+a_(1)x+a_(2)x^(2)+...+a_(50)x^(50)` for all `x in R`, then `(a_(2))/(a_(0))` is equal to

Options

A. 12.5
B. 12
C. 12.75
D. 12.25

Answer

To solve the problem, we need to find the ratio \( \frac{a_2}{a_0} \) from the expression \( (x + 10)^{50} + (x - 10)^{50} \).

### Step-by-Step Solution:

1. **Understand the Expression**:
We have the expression \( (x + 10)^{50} + (x - 10)^{50} \). This can be expanded using the binomial theorem.

2. **Expand Using Binomial Theorem**:
The binomial expansion for \( (x + 10)^{50} \) is:
\[ (x + 10)^{50} = \sum_{r=0}^{50} \binom{50}{r} x^{50-r} 10^r \]
Similarly, for \( (x - 10)^{50} \):
\[ (x - 10)^{50} = \sum_{r=0}^{50} \binom{50}{r} x^{50-r} (-10)^r \]

3. **Combine the Expansions**:
Adding both expansions:
\[ (x + 10)^{50} + (x - 10)^{50} = \sum_{r=0}^{50} \binom{50}{r} x^{50-r} (10^r + (-10)^r) \]
Notice that \( 10^r + (-10)^r \) is zero for odd \( r \) and \( 2 \cdot 10^r \) for even \( r \).

4. **Identify Coefficients**:
Therefore, the combined expression simplifies to:
\[ = \sum_{k=0}^{25} \binom{50}{2k} x^{50-2k} (2 \cdot 10^{2k}) \]
Here, \( a_0 \) is the constant term (when \( k = 25 \)) and \( a_2 \) is the coefficient of \( x^2 \) (when \( k = 24 \)).

5. **Calculate \( a_0 \)**:
For \( k = 25 \):
\[ a_0 = \binom{50}{50} \cdot 2 \cdot 10^{50} = 2 \cdot 10^{50} \]

6. **Calculate \( a_2 \)**:
For \( k = 24 \):
\[ a_2 = \binom{50}{48} \cdot 2 \cdot 10^{48} = \binom{50}{2} \cdot 2 \cdot 10^{48} \]
Here, \( \binom{50}{48} = \binom{50}{2} = \frac{50 \cdot 49}{2} = 1225 \).

7. **Final Values**:
Thus,
\[ a_2 = 1225 \cdot 2 \cdot 10^{48} = 2450 \cdot 10^{48} \]
and
\[ a_0 = 2 \cdot 10^{50} \]

8. **Calculate \( \frac{a_2}{a_0} \)**:
\[ \frac{a_2}{a_0} = \frac{2450 \cdot 10^{48}}{2 \cdot 10^{50}} = \frac{2450}{2} \cdot \frac{10^{48}}{10^{50}} = \frac{2450}{2} \cdot 10^{-2} = 1225 \cdot 10^{-2} = 12.25 \]

### Final Answer:
\[ \frac{a_2}{a_0} = 12.25 \]