The sum `""^(40)C_(0) + ""^(40)C_(1)+""^(40)C_(2)+…+""^(40)C_(20) `is equal to

Options

A. `2^(40) + (40!)/((20!)^(2))`
B. `2^(39)+(1)/(2) (40!)/((20!)^(2))`
C. `2^(39) + ""^(40)C_(20)`
D. none of these

Answer

To find the sum \( \binom{40}{0} + \binom{40}{1} + \binom{40}{2} + \ldots + \binom{40}{20} \), we can use the properties of binomial coefficients and the Binomial Theorem.

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**:
The Binomial Theorem states that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For \( n = 40 \), we have:
\[ (1 + x)^{40} = \sum_{k=0}^{40} \binom{40}{k} x^k \]

2. **Setting \( x = 1 \)**:
By substituting \( x = 1 \) into the binomial expansion, we get:
\[ (1 + 1)^{40} = 2^{40} = \sum_{k=0}^{40} \binom{40}{k} \]
This means:
\[ \sum_{k=0}^{40} \binom{40}{k} = 2^{40} \]

3. **Using Symmetry of Binomial Coefficients**:
We know that \( \binom{n}{k} = \binom{n}{n-k} \). Therefore:
\[ \binom{40}{0} = \binom{40}{40}, \quad \binom{40}{1} = \binom{40}{39}, \quad \ldots, \quad \binom{40}{20} = \binom{40}{20} \]
This symmetry implies that the sum of the first half of the coefficients is equal to the sum of the second half.

4. **Splitting the Total Sum**:
Since there are 41 terms in total (from \( k = 0 \) to \( k = 40 \)), we can express the total sum as:
\[ \sum_{k=0}^{40} \binom{40}{k} = \sum_{k=0}^{20} \binom{40}{k} + \sum_{k=21}^{40} \binom{40}{k} \]
Due to the symmetry, we have:
\[ \sum_{k=21}^{40} \binom{40}{k} = \sum_{k=0}^{19} \binom{40}{k} \]
Thus:
\[ \sum_{k=0}^{20} \binom{40}{k} + \sum_{k=0}^{20} \binom{40}{k} = 2 \sum_{k=0}^{20} \binom{40}{k} = 2^{40} \]

5. **Finding the Required Sum**:
Therefore, we can conclude:
\[ 2 \sum_{k=0}^{20} \binom{40}{k} = 2^{40} \]
Dividing both sides by 2 gives:
\[ \sum_{k=0}^{20} \binom{40}{k} = 2^{39} \]

### Final Answer:
The sum \( \binom{40}{0} + \binom{40}{1} + \binom{40}{2} + \ldots + \binom{40}{20} \) is equal to \( 2^{39} \).

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