To solve the problem \(\sum_{k=0}^{10} {}^{20}C_k\), we need to understand that it involves the binomial coefficients. The binomial theorem states that:
\((a + b)^n = \sum_{k=0}^{n} {}^{n}C_k \cdot a^{n-k} \cdot b^{k}\)
By choosing \(a = 1\) and \(b = 1\), the theorem simplifies to:
\((1 + 1)^{20} = \sum_{k=0}^{20} {}^{20}C_k = 2^{20}\)
Similarly, choosing \(a = 1\) and \(b = -1\), it gives:
\((1 - 1)^{20} = \sum_{k=0}^{20} {}^{20}C_k \cdot (-1)^k = 0\)
Which means:
\(\sum_{k=0}^{20} {}^{20}C_k \cdot (-1)^k = {}^{20}C_0 - {}^{20}C_1 + {}^{20}C_2 - \ldots + {}^{20}C_{20} = 0\)
Adding the equations:
We get:
\(2 \times \sum_{k=0}^{10} {}^{20}C_{2k} = 2^{20}\)
\(\sum_{k=0}^{10} {}^{20}C_{2k} = 2^{19}\)
Now, considering the symmetry and properties of binomial coefficients, adding up from \(k=0\) to \(k=10\) and using the identity:
\(\sum_{k=0}^{n} {}^{n}C_k = 2^{n-1} + \frac{1}{2} \cdot {}^{n}C_{\frac{n}{2}}\)
Since \(n = 20\) is even, \(k=10\) is the middle term:
The result is:
\(\sum_{k=0}^{10} {}^{20}C_k = 2^{19} + \frac{1}{2} \cdot {}^{20}C_{10}\)
Therefore, the correct answer is \(2^{19} + \frac{1}{2} \cdot {}^{20}C_{10}\)