The sum of the series `""^20C_0+""^20C_1+""^20C_2+...+""^20C_9` is =

Options

A. `2^20`
B. `2^19`
C. `2^19+1/2*""^20C_10`
D. `2^19-1/2*""^20C_10`

Answer

To solve the problem, we need to find the sum of the series:

\[ \sum_{k=0}^{9} \binom{20}{k} \]

### Step 1: Understanding the Binomial Theorem
The Binomial Theorem states that:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

For \( n = 20 \) and \( x = 1 \), we have:

\[ (1 + 1)^{20} = 2^{20} = \sum_{k=0}^{20} \binom{20}{k} \]

This means that the sum of all the binomial coefficients from \( k = 0 \) to \( k = 20 \) is \( 2^{20} \).

### Step 2: Splitting the Sum
We can split the sum of the binomial coefficients into two parts:

\[ \sum_{k=0}^{20} \binom{20}{k} = \sum_{k=0}^{9} \binom{20}{k} + \sum_{k=10}^{20} \binom{20}{k} \]

### Step 3: Using the Symmetry Property
We can use the property of binomial coefficients, which states that:

\[ \binom{n}{r} = \binom{n}{n-r} \]

For our case, this gives us:

\[ \sum_{k=10}^{20} \binom{20}{k} = \sum_{k=0}^{10} \binom{20}{k} \]

Thus, we can rewrite the total sum as:

\[ 2^{20} = \sum_{k=0}^{9} \binom{20}{k} + \sum_{k=10}^{20} \binom{20}{k} \]

### Step 4: Combining the Sums
Notice that:

\[ \sum_{k=10}^{20} \binom{20}{k} = \sum_{k=0}^{10} \binom{20}{k} \]

This means:

\[ 2^{20} = \sum_{k=0}^{9} \binom{20}{k} + \sum_{k=0}^{10} \binom{20}{k} \]

### Step 5: Isolating the Desired Sum
Now, we can express the sum we want to find:

\[ \sum_{k=0}^{9} \binom{20}{k} = \frac{2^{20}}{2} - \frac{1}{2} \binom{20}{10} \]

This simplifies to:

\[ \sum_{k=0}^{9} \binom{20}{k} = 2^{19} - \frac{1}{2} \binom{20}{10} \]

### Step 6: Final Calculation
Now we need to calculate \( 2^{19} \) and \( \binom{20}{10} \):

- \( 2^{20} = 1048576 \) so \( 2^{19} = 524288 \)
- The value of \( \binom{20}{10} = 184756 \)

Putting it all together:

\[ \sum_{k=0}^{9} \binom{20}{k} = 524288 - \frac{184756}{2} \]

Calculating \( \frac{184756}{2} = 92378 \):

\[ \sum_{k=0}^{9} \binom{20}{k} = 524288 - 92378 = 431910 \]

Thus, the final answer is:

\[ \sum_{k=0}^{9} \binom{20}{k} = 431910 \]