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 evaluate the expression \((x + 10)^{50} + (x - 10)^{50}\) and find the coefficients \(a_0\) and \(a_2\) in the polynomial expansion. We will then compute the ratio \(\frac{a_2}{a_0}\).

### Step-by-Step Solution:

1. **Expand the Expression**:
We start with the expression:
\[ (x + 10)^{50} + (x - 10)^{50} \]
This can be expanded using the Binomial Theorem:
\[ (x + 10)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^{50-k} (10)^k \]
\[ (x - 10)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^{50-k} (-10)^k \]

2. **Combine the Expansions**:
Adding these two expansions together, we notice that the odd powers of \(10\) will cancel out:
\[ (x + 10)^{50} + (x - 10)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^{50-k} (10^k + (-10)^k) \]
The terms where \(k\) is odd will cancel out, leaving only the even powers:
\[ = 2 \sum_{k \text{ even}} \binom{50}{k} x^{50-k} 10^k \]

3. **Identify Coefficients**:
The coefficients \(a_0\) and \(a_2\) correspond to:
- \(a_0\) is the coefficient of \(x^{50}\) (when \(k=0\)):
\[ a_0 = 2 \cdot \binom{50}{0} \cdot 10^0 = 2 \]
- \(a_2\) is the coefficient of \(x^{48}\) (when \(k=2\)):
\[ a_2 = 2 \cdot \binom{50}{2} \cdot 10^2 \]
We calculate \(\binom{50}{2}\):
\[ \binom{50}{2} = \frac{50 \cdot 49}{2} = 1225 \]
Thus,
\[ a_2 = 2 \cdot 1225 \cdot 100 = 245000 \]

4. **Calculate the Ratio**:
Now we compute the ratio \(\frac{a_2}{a_0}\):
\[ \frac{a_2}{a_0} = \frac{245000}{2} = 122500 \]

5. **Final Calculation**:
To express this in terms of the original problem, we note that:
\[ \frac{a_2}{a_0} = \frac{245000}{2} = 122500 \]

### Conclusion:
Thus, the final answer is:
\[ \frac{a_2}{a_0} = 122500 \]