If sum of the coefficients in the expansion of `(x + y)^(n)` is 2048, then the greatest coefficient in the expansion is:

Options

A. `""^(10)C_(6)`
B. `""^(11)C_(6)`
C. `""^(10)C_(7)`
D. `""^(12)C_(6)`

Answer

To solve the problem step-by-step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Problem
We need to find the greatest coefficient in the expansion of \((x + y)^n\) given that the sum of the coefficients is 2048.

### Step 2: Find the Value of \(n\)
The sum of the coefficients in the expansion of \((x + y)^n\) can be found by substituting \(x = 1\) and \(y = 1\):
\[ (1 + 1)^n = 2^n \]
According to the problem, this sum is equal to 2048:
\[ 2^n = 2048 \]
Now, we can express 2048 as a power of 2:
\[ 2048 = 2^{11} \]
Thus, we have:
\[ n = 11 \]

### Step 3: Identify the Greatest Coefficient
In the expansion of \((x + y)^n\), the coefficients are given by the binomial coefficients \(\binom{n}{r}\) where \(r\) ranges from 0 to \(n\). The greatest coefficient occurs at the middle terms when \(n\) is even or at the two middle terms when \(n\) is odd.

Since \(n = 11\) (which is odd), the greatest coefficients will be at:
- The 6th term: \(\binom{11}{5}\)
- The 7th term: \(\binom{11}{6}\)

### Step 4: Calculate the Coefficients
Now we will calculate the coefficients:
1. For the 6th term:
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} \]
Calculating this gives:
\[ \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = \frac{55440}{120} = 462 \]

2. For the 7th term:
\[ \binom{11}{6} = \frac{11!}{6!(11-6)!} = \frac{11!}{6!5!} \]
This is the same as \(\binom{11}{5}\) due to the symmetry of binomial coefficients:
\[ \binom{11}{6} = 462 \]

### Step 5: Conclusion
Thus, the greatest coefficient in the expansion of \((x + y)^{11}\) is:
\[ \text{Greatest Coefficient} = 462 \]