The sum of the last eitht coefficients in the
expansion of `(1 + x)^(15)` , is

Options

A. `2^(16)`
B. `2^(15)`
C. `2^(14)`
D. none of these

Explanation

To find the sum of the last eight coefficients in the expansion of \((1 + x)^{15}\), we can follow these steps:

### Step 1: Identify the coefficients
The coefficients in the expansion of \((1 + x)^{15}\) are given by the binomial coefficients \(\binom{15}{k}\) for \(k = 0, 1, 2, \ldots, 15\). The last eight coefficients correspond to \(k = 8\) to \(k = 15\), which are:
\[ \binom{15}{8}, \binom{15}{9}, \binom{15}{10}, \binom{15}{11}, \binom{15}{12}, \binom{15}{13}, \binom{15}{14}, \binom{15}{15} \]

### Step 2: Write the sum of the coefficients
The sum of the last eight coefficients can be expressed as:
\[ S_n = \binom{15}{8} + \binom{15}{9} + \binom{15}{10} + \binom{15}{11} + \binom{15}{12} + \binom{15}{13} + \binom{15}{14} + \binom{15}{15} \]

### Step 3: Use the property of binomial coefficients
We can use the property of binomial coefficients that states:
\[ \binom{n}{r} = \binom{n}{n-r} \]
This means we can rewrite the coefficients:
\[ \binom{15}{8} = \binom{15}{7}, \quad \binom{15}{9} = \binom{15}{6}, \quad \binom{15}{10} = \binom{15}{5}, \quad \binom{15}{11} = \binom{15}{4}, \quad \binom{15}{12} = \binom{15}{3}, \quad \binom{15}{13} = \binom{15}{2}, \quad \binom{15}{14} = \binom{15}{1}, \quad \binom{15}{15} = \binom{15}{0} \]

### Step 4: Rewrite the sum
Thus, we can rewrite the sum \(S_n\) as:
\[ S_n = \binom{15}{7} + \binom{15}{6} + \binom{15}{5} + \binom{15}{4} + \binom{15}{3} + \binom{15}{2} + \binom{15}{1} + \binom{15}{0} \]

### Step 5: Use the binomial theorem
The sum of all coefficients in the expansion of \((1 + x)^{15}\) is given by:
\[ (1 + 1)^{15} = 2^{15} \]
This includes all coefficients from \(k = 0\) to \(k = 15\).

### Step 6: Relate the sums
Since the coefficients are symmetric, the sum of the first eight coefficients (from \(k = 0\) to \(k = 7\)) is equal to the sum of the last eight coefficients (from \(k = 8\) to \(k = 15\)):
\[ \text{Sum of first 8 coefficients} = \text{Sum of last 8 coefficients} = \frac{1}{2} \cdot 2^{15} = 2^{14} \]

### Final Answer
Thus, the sum of the last eight coefficients in the expansion of \((1 + x)^{15}\) is:
\[ \boxed{16384} \quad (\text{which is } 2^{14}) \]