If `Sigma_(i=1)^(20) ((""^(20)C_(i-1))/(""^(20)C_(i)+""^(20)C_(i-1)))^(3)=(k)/(21)`, then k equals

Options

A. 200
B. 50
C. 100
D. 400

Answer

To solve the problem, we need to evaluate the summation given in the question and find the value of \( k \).

### Step-by-Step Solution

1. **Understanding the Summation**:
We are given:
\[ \sum_{i=1}^{20} \left( \frac{\binom{20}{i-1}}{\binom{20}{i} + \binom{20}{i-1}} \right)^3 = \frac{k}{21} \]

2. **Simplifying the Fraction**:
The term inside the summation can be simplified:
\[ \frac{\binom{20}{i-1}}{\binom{20}{i} + \binom{20}{i-1}} \]
We know that:
\[ \binom{20}{i} = \frac{20!}{i!(20-i)!} \quad \text{and} \quad \binom{20}{i-1} = \frac{20!}{(i-1)!(20-i+1)!} \]
Thus, we can rewrite the fraction:
\[ \frac{\binom{20}{i-1}}{\binom{20}{i} + \binom{20}{i-1}} = \frac{\frac{20!}{(i-1)!(20-i+1)!}}{\frac{20!}{i!(20-i)!} + \frac{20!}{(i-1)!(20-i+1)!}} \]
After canceling \( 20! \) from the numerator and denominator, we get:
\[ = \frac{1}{\frac{1}{i} + \frac{1}{21-i}} = \frac{i(21-i)}{21} \]

3. **Cubing the Result**:
Now we cube the result:
\[ \left( \frac{i(21-i)}{21} \right)^3 = \frac{i^3(21-i)^3}{21^3} \]

4. **Substituting Back into the Summation**:
The summation now becomes:
\[ \sum_{i=1}^{20} \left( \frac{i(21-i)}{21} \right)^3 = \frac{1}{21^3} \sum_{i=1}^{20} i^3 (21-i)^3 \]

5. **Using the Identity for \( i^3 \)**:
We can use the identity for the sum of cubes:
\[ \sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
For \( n = 20 \):
\[ \sum_{i=1}^{20} i^3 = \left( \frac{20 \cdot 21}{2} \right)^2 = 10^2 \cdot 21^2 = 100 \cdot 441 = 44100 \]

6. **Final Calculation**:
Now substituting back, we have:
\[ \sum_{i=1}^{20} i^3 = 44100 \]
Therefore:
\[ \frac{1}{21^3} \cdot 44100 = \frac{44100}{9261} = \frac{44100}{21^3} \]
Setting this equal to \( \frac{k}{21} \):
\[ \frac{44100}{21^3} = \frac{k}{21} \]
Multiplying both sides by \( 21^2 \):
\[ k = \frac{44100}{21^2} = \frac{44100}{441} = 100 \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{100} \]