If the sum of the coefficients in the expansion of `(p+q)^(n)` is 1024, then the greatest coefficient in the expansion is

Options

A. `""^(10) C_5`
B. `""^(10) C_4`
C. `""^(10) C_(2)`
D. None of these

Answer

To solve the problem step by step, we need to find the greatest coefficient in the expansion of \((p + q)^n\) given that the sum of the coefficients is 1024.

### Step 1: Understanding the Sum of Coefficients
The sum of the coefficients in the expansion of \((p + q)^n\) can be found by substituting \(p = 1\) and \(q = 1\):
\[ (1 + 1)^n = 2^n \]
Given that this sum is equal to 1024, we can set up the equation:
\[ 2^n = 1024 \]

### Step 2: Solving for \(n\)
Next, we need to express 1024 as a power of 2:
\[ 1024 = 2^{10} \]
Thus, we can equate the exponents:
\[ n = 10 \]

### Step 3: Finding the Greatest Coefficient
In the expansion of \((p + q)^{10}\), the coefficients are given by the binomial coefficients \(\binom{10}{r}\) for \(r = 0, 1, 2, \ldots, 10\). The greatest coefficient occurs at \(r = \frac{n}{2}\) when \(n\) is even.

Since \(n = 10\) is even, we find:
\[ r = \frac{10}{2} = 5 \]

### Step 4: Calculating the Greatest Coefficient
The greatest coefficient is thus:
\[ \binom{10}{5} \]

### Step 5: Computing \(\binom{10}{5}\)
We can calculate \(\binom{10}{5}\) using the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Substituting \(n = 10\) and \(r = 5\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30240}{120} = 252 \]

### Final Answer
Thus, the greatest coefficient in the expansion of \((p + q)^{10}\) is:
\[ \boxed{252} \]