Sum of the series `sum_(k=0)^(n)((2n)!)/((k!)^(2)(n-k!)^(2))` equals

Options

A. `n^(n)+""^(2n)C_(n)`
B. `(""^(2n)C_(n))^(2)`
C. `""^(2n)C_(n)`
D. `n^(n)`

Answer

To solve the problem, we need to evaluate the sum of the series:

\[ S = \sum_{k=0}^{n} \frac{(2n)!}{(k!)^2 (n-k)!^2} \]

### Step 1: Rewrite the series
We can rewrite the series using the binomial coefficient notation. The term \(\frac{(2n)!}{(k!)^2 (n-k)!^2}\) can be recognized as the number of ways to choose \(k\) items from \(2n\) items, where we are selecting \(k\) items in two groups of \(n\).

Thus, we can express the sum as:

\[ S = \sum_{k=0}^{n} \binom{2n}{k} \binom{n}{k} \]

### Step 2: Apply the Vandermonde Identity
We can apply the Vandermonde identity, which states that:

\[ \sum_{k=0}^{r} \binom{x}{k} \binom{y}{r-k} = \binom{x+y}{r} \]

In our case, we can set \(x = n\), \(y = n\), and \(r = n\):

\[ S = \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k} = \binom{2n}{n} \]

### Step 3: Final Result
Thus, the sum of the series is:

\[ S = \binom{2n}{n} \]

### Conclusion
The final answer is:

\[ \sum_{k=0}^{n} \frac{(2n)!}{(k!)^2 (n-k)!^2} = \binom{2n}{n} \]