What is the value of $^{23}C_0 + {}^{23}C_2 + {}^{23}C_4 + \dots + {}^{23}C_{22}$ ?
Show 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 \(2^{22}\).
• Step 1 : Understanding the Question:
We are asked to find the sum of alternate binomial coefficients starting from \({}^{23}C_0\) up to \({}^{23}C_{22}\).
These are the binomial coefficients with even indices for \(n = 23\).
• Step 2 : Key Formula or Approach:
According to the binomial theorem:
\[ (1 + x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n \]
By substituting \(x = 1\) and \(x = -1\)., we get two equations.
Adding these two equations eliminates all the odd-indexed terms, leaving only twice the sum of the even-indexed terms.
• Step 3 : Detailed Explanation:
Let us write down the expansion for \(n = 23\):
\[ (1 + x)^{23} = {}^{23}C_0 + {}^{23}C_1 x + {}^{23}C_2 x^2 + \dots + {}^{23}C_{23} x^{23} \]
Substitute \(x = 1\):
\[ 2^{23} = {}^{23}C_0 + {}^{23}C_1 + {}^{23}C_2 + \dots + {}^{23}C_{23} \quad \text{--- (Equation 1)} \]
Substitute \(x = -1\):
\[ 0 = {}^{23}C_0 - {}^{23}C_1 + {}^{23}C_2 - {}^{23}C_3 + \dots - {}^{23}C_{23} \quad \text{--- (Equation 2)} \]
Now, add Equation 1 and Equation 2:
\[ 2^{23} + 0 = 2 \left( {}^{23}C_0 + {}^{23}C_2 + {}^{23}C_4 + \dots + {}^{23}C_{22} \right) \]
Dividing both sides by 2:
\[ {}^{23}C_0 + {}^{23}C_2 + {}^{23}C_4 + \dots + {}^{23}C_{22} = \frac{2^{23}}{2} = 2^{22} \]
• Step 4 : Final Answer:
The value of the sum is \(2^{22}\).
This corresponds to option (A).
