Question

$\sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k}$

• A. $\binom{2n}{n}$
• B. $4^n$
• C. $2^{2n}$
• D. $\binom{4n}{2n}$

Answer 1

Answer: (B)

$4^n$

Answer 2

Let the given summation be $S_n$. We are asked to find $S_n = \sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k}$. We use the generating function for the central binomial coefficients, which is $f(x) = \sum_{m=0}^{\infty} \binom{2m}{m} x^m = \frac{1}{\sqrt{1-4x}}$. The given sum $S_n$ is the coefficient of $x^n$ in the product of $f(x)$ with itself, i.e., $f(x) \cdot f(x)$. $f(x)^2 = \left(\frac{1}{\sqrt{1-4x}}\right)^2 = \frac{1}{1-4x}$. Expanding $\frac{1}{1-4x}$ as a geometric series, we get: $\frac{1}{1-4x} = \sum_{n=0}^{\infty} (4x)^n = \sum_{n=0}^{\infty} 4^n x^n$. The coefficient of $x^n$ in this expansion is $4^n$. Therefore, $S_n = 4^n$.