IF `C_k` stands for `""^n C_k` then `sum_(k=1)^n ((kC_k)/(C_k+C_(n-k)))^2` is equal to

Options

A. `(n(n+1)(2n+1))/24`
B. `(n^2(n+1))^2/8`
C. `(n^3(n+2))/16`
D. `(n^2(2n+1))/12`

Answer

To solve the problem, we need to evaluate the expression:

\[ S = \sum_{k=1}^{n} \left( \frac{k C_k}{C_k + C_{n-k}} \right)^2 \]

where \( C_k \) stands for \( \binom{n}{k} \).

### Step 1: Rewrite the Expression
First, we recognize that \( C_k = \binom{n}{k} \) and \( C_{n-k} = \binom{n}{n-k} \). By the symmetry property of binomial coefficients, we have:

\[ C_k = C_{n-k} \]

Thus, we can rewrite the denominator:

\[ C_k + C_{n-k} = C_k + C_k = 2C_k \]

### Step 2: Substitute Back into the Summation
Now we can substitute this back into our summation:

\[ S = \sum_{k=1}^{n} \left( \frac{k C_k}{2 C_k} \right)^2 = \sum_{k=1}^{n} \left( \frac{k}{2} \right)^2 \]

### Step 3: Factor Out the Constant
Since \( \left( \frac{k}{2} \right)^2 = \frac{k^2}{4} \), we can factor out \( \frac{1}{4} \):

\[ S = \frac{1}{4} \sum_{k=1}^{n} k^2 \]

### Step 4: Use the Formula for the Sum of Squares
The formula for the sum of the squares of the first \( n \) natural numbers is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 5: Substitute the Formula into the Expression
Now, substituting this formula into our expression for \( S \):

\[ S = \frac{1}{4} \cdot \frac{n(n+1)(2n+1)}{6} \]

### Step 6: Simplify the Expression
Now, we simplify:

\[ S = \frac{n(n+1)(2n+1)}{24} \]

### Final Result
Thus, the final result is:

\[ S = \frac{n(n+1)(2n+1)}{24} \]