Question:

${}^{21}C_1 + {}^{21}C_2 + \dots + {}^{21}C_{10} =$

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The sum of the first half of binomial coefficients for odd $n$ is $2^{n-1}$.
Updated On: Apr 28, 2026
  • $2^{20}$
  • $2^{21}$
  • $2^{21} - 1$
  • $2^{21} - 2$
  • $2^{20} - 1$
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The Correct Option is

Solution and Explanation

Step 1: Concept
Sum of all coefficients: $\sum_{r=0}^{21} {}^{21}C_r = 2^{21}$. Also, ${}^{21}C_r = {}^{21}C_{21-r}$.

Step 2: Analysis

$2^{21} = ({}^{21}C_0 + \dots + {}^{21}C_{10}) + ({}^{21}C_{11} + \dots + {}^{21}C_{21})$. Since the two halves are equal: $2 \times ({}^{21}C_0 + \dots + {}^{21}C_{10}) = 2^{21}$. ${}^{21}C_0 + {}^{21}C_1 + \dots + {}^{21}C_{10} = 2^{20}$.

Step 3: Calculation

The required sum is $({}^{21}C_0 + {}^{21}C_1 + \dots + {}^{21}C_{10}) - {}^{21}C_0$. Sum $= 2^{20} - 1$. Final Answer: (E)
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