Question

What is the value of $^{23}C_0 + {}^{23}C_2 + {}^{23}C_4 + \dots + {}^{23}C_{22}$ ?

• A. $2^{22}$
• B. $2^{22} - 1$
• C. $2^{23} + 1$
• D. $2^{23}$

Hint:

For any odd power \(n\)., the sum of even-indexed binomial coefficients is always exactly equal to half of the total sum of binomial coefficients:
\[ \text{Sum} = \frac{2^n}{2} = 2^{n-1} \] With \(n = 23\)., the answer is directly \(2^{22}\).

Answer 1

The Correct Option is A

To solve the problem of finding the value of \(^{23}C_0 + {}^{23}C_2 + {}^{23}C_4 + \dots + {}^{23}C_{22}\), we first need to understand that this is a problem of binomial coefficients. Specifically, this sum represents the sum of every alternate binomial coefficient from the binomial expansion of \((1 + x)^{23}\) evaluated at \(x = 1\) and \(x = -1\).

Step-by-Step Solution:

The alternating sum of the binomial coefficients can be written as the following two separate sums:

1. \(S_{\text{odd}} = \sum_{\text{odd } k} {}^{23}C_k\) and \(S_{\text{even}} = \sum_{\text{even } k} {}^{23}C_k\).

The binomial theorem tells us:

1. \((1 + x)^{23} = \sum_{k=0}^{23} {}^{23}C_k x^k\).

Setting \(x = 1\) gives:

1. \(2^{23} = \sum_{k=0}^{23} {}^{23}C_k\).

This equation sums all the binomial coefficients.

Now, setting \(x = -1\) gives:

1. \(0 = \sum_{k=0}^{23} {}^{23}C_k (-1)^k = \sum_{\text{even } k} {}^{23}C_k - \sum_{\text{odd } k} {}^{23}C_k\).

From the above equation, we find that:

1. \(S_{\text{even}} - S_{\text{odd}} = 0\).

This implies:

1. \(S_{\text{even}} = S_{\text{odd}}\).

Thus, the sum of all coefficients from \(x^0\) to \(x^{22}\) (i.e., the terms we are given) can be expressed as:

1. \(S_{\text{even}} = \frac{2^{23}}{2} = 2^{22}\).

Therefore, the value of the given sum \(\displaystyle{}^{23}C_0 + {}^{23}C_2 + {}^{23}C_4 + \ldots + {}^{23}C_{22} \) is \(2^{22}\).