Question

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

• A. $2^{20}$
• B. $2^{21}$
• C. $2^{21} - 1$
• D. $2^{21} - 2$
• E. $2^{20} - 1$

Hint:

The sum of the first half of binomial coefficients for odd $n$ is $2^{n-1}$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We are asked to find the sum of the first 10 binomial coefficients for n=21. This is not the full sum, but a partial sum. We can use the properties of binomial coefficients to solve this.
Step 2: Key Formula or Approach:
We will use two key properties:
1. The sum of all binomial coefficients: \( \sum_{r=0}^{n} \binom{n}{r} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n \).
2. The symmetry property: \( \binom{n}{r} = \binom{n}{n-r} \).
Step 3: Detailed Explanation:
Let's write out the full sum of binomial coefficients for n=21:
\[ ^{21}C_0 + ^{21}C_1 + ^{21}C_2 + \dots + ^{21}C_{10} + ^{21}C_{11} + \dots + ^{21}C_{20} + ^{21}C_{21} = 2^{21} \] Let the sum we want to find be S.
\[ S = ^{21}C_1 + ^{21}C_2 + \dots + ^{21}C_{10} \] Now, let's use the symmetry property \( \binom{21}{r} = \binom{21}{21-r} \):
\( ^{21}C_{20} = ^{21}C_{21-20} = ^{21}C_1 \)
\( ^{21}C_{19} = ^{21}C_{21-19} = ^{21}C_2 \)
...
\( ^{21}C_{11} = ^{21}C_{21-11} = ^{21}C_{10} \)
So, the sum \( ^{21}C_{11} + ^{21}C_{12} + \dots + ^{21}C_{20} \) is equal to \( ^{21}C_{10} + ^{21}C_9 + \dots + ^{21}C_1 \), which is also S.
Now we can rewrite the full sum:
\[ ^{21}C_0 + ( ^{21}C_1 + \dots + ^{21}C_{10}) + ( ^{21}C_{11} + \dots + ^{21}C_{20}) + ^{21}C_{21} = 2^{21} \] Substitute S into the equation:
\[ ^{21}C_0 + S + S + ^{21}C_{21} = 2^{21} \] We know that \( ^{21}C_0 = 1 \) and \( ^{21}C_{21} = 1 \).
\[ 1 + 2S + 1 = 2^{21} \] \[ 2 + 2S = 2^{21} \] Now, we solve for S:
\[ 2S = 2^{21} - 2 \] Divide the entire equation by 2:
\[ S = \frac{2^{21}}{2} - \frac{2}{2} \] \[ S = 2^{20} - 1 \] Step 4: Final Answer:
The value of the sum is \( 2^{20} - 1 \).