Question:
\(\sum_{k=0}^{10} {}^{20}C_k\) is equal to
\(\sum_{k=0}^{10} {}^{20}C_k\) is equal to
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Use \(\sum_{k=0}^n \binom{n}{k} = 2^n\) and symmetry \(\binom{n}{k} = \binom{n}{n-k}\).
Updated On: Apr 23, 2026
- \(2^{19} + \frac{1}{2} {}^{20}C_{10}\)
- \(2^{19}\)
- \({}^{20}C_{10}\)
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \sum_{k=0}^{20} {}^{20}C_k = 2^{20} \]
Step 2: Calculation / Simplification}
By symmetry: \(\sum_{k=0}^{9} {}^{20}C_k = \sum_{k=11}^{20} {}^{20}C_k\)
\[ \sum_{k=0}^{10} {}^{20}C_k + \sum_{k=11}^{20} {}^{20}C_k = 2^{20} \]
\[ \sum_{k=0}^{10} {}^{20}C_k + \left(\sum_{k=0}^{9} {}^{20}C_k\right) = 2^{20} \]
\[ 2\sum_{k=0}^{10} {}^{20}C_k - {}^{20}C_{10} = 2^{20} \]
\[ \sum_{k=0}^{10} {}^{20}C_k = 2^{19} + \frac{1}{2} {}^{20}C_{10} \]
Step 3: Final Answer
\[ 2^{19} + \frac{1}{2} {}^{20}C_{10} \]
\[ \sum_{k=0}^{20} {}^{20}C_k = 2^{20} \]
Step 2: Calculation / Simplification}
By symmetry: \(\sum_{k=0}^{9} {}^{20}C_k = \sum_{k=11}^{20} {}^{20}C_k\)
\[ \sum_{k=0}^{10} {}^{20}C_k + \sum_{k=11}^{20} {}^{20}C_k = 2^{20} \]
\[ \sum_{k=0}^{10} {}^{20}C_k + \left(\sum_{k=0}^{9} {}^{20}C_k\right) = 2^{20} \]
\[ 2\sum_{k=0}^{10} {}^{20}C_k - {}^{20}C_{10} = 2^{20} \]
\[ \sum_{k=0}^{10} {}^{20}C_k = 2^{19} + \frac{1}{2} {}^{20}C_{10} \]
Step 3: Final Answer
\[ 2^{19} + \frac{1}{2} {}^{20}C_{10} \]
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