The sum of last ten coefficients in the expansion of `(1+x)^(19)` when expanded in ascending powers of x is

Options

A. `2^(19)`
B. `2^(18)`
C. `2^(18)-""^(19)C_(1)`
D. none

Answer

To find the sum of the last ten coefficients in the expansion of \( (1+x)^{19} \), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \( (1+x)^n \) is given by:
\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For \( n = 19 \), the expansion becomes:
\[ (1+x)^{19} = \binom{19}{0} + \binom{19}{1}x + \binom{19}{2}x^2 + \ldots + \binom{19}{19}x^{19} \]

### Step 2: Identify the Last Ten Coefficients
The last ten coefficients correspond to the terms from \( \binom{19}{10} \) to \( \binom{19}{19} \). We need to find the sum:
\[ \binom{19}{10} + \binom{19}{11} + \binom{19}{12} + \binom{19}{13} + \binom{19}{14} + \binom{19}{15} + \binom{19}{16} + \binom{19}{17} + \binom{19}{18} + \binom{19}{19} \]

### Step 3: Use the Binomial Theorem Property
From the binomial theorem, we know that:
\[ (1+x)^n \text{ evaluated at } x=1 \text{ gives the sum of all coefficients.} \]
Thus,
\[ (1+1)^{19} = 2^{19} = \sum_{k=0}^{19} \binom{19}{k} \]

### Step 4: Relate the Coefficients
The sum of the coefficients from \( \binom{19}{0} \) to \( \binom{19}{9} \) can be expressed as:
\[ \sum_{k=0}^{9} \binom{19}{k} \]
Using the symmetry property of binomial coefficients:
\[ \binom{19}{k} = \binom{19}{19-k} \]
we can relate:
\[ \sum_{k=0}^{9} \binom{19}{k} = \sum_{k=10}^{19} \binom{19}{k} \]
Thus, the sum of the last ten coefficients is half of the total sum of coefficients:
\[ \sum_{k=10}^{19} \binom{19}{k} = \frac{1}{2} \cdot 2^{19} = 2^{18} \]

### Step 5: Final Result
Therefore, the sum of the last ten coefficients in the expansion of \( (1+x)^{19} \) is:
\[ \boxed{262144} \]