If `(1+x)(1+x^2)(1+x^4)….(1+x^(128))=Sigma_(r=0)^(n) x^r`, then n is equal is

Options

A. 256
B. 255
C. 254
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \((1+x)(1+x^2)(1+x^4)\cdots(1+x^{128})\) and find the value of \(n\) such that it equals \(\Sigma_{r=0}^{n} x^r\).

### Step-by-Step Solution:

1. **Understand the Expression**:
The expression given is \((1+x)(1+x^2)(1+x^4)\cdots(1+x^{128})\). Each term can be expanded, and we need to find the total degree of \(x\) in the expansion.

2. **Identify the Terms**:
The terms in the product are \(1 + x^k\) where \(k\) takes values from \(1, 2, 4, 8, 16, 32, 64, 128\). This means we have the powers of \(x\) as \(1, 2, 4, 8, 16, 32, 64, 128\).

3. **Count the Terms**:
The powers of \(x\) can be expressed as \(2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7\). This gives us a total of \(8\) terms (from \(2^0\) to \(2^7\)).

4. **Calculate the Maximum Degree**:
The maximum degree of \(x\) in the expansion will be the sum of all the powers:
\[ 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255 \]
This is a geometric series where the first term \(a = 1\) and the common ratio \(r = 2\). The sum of the first \(n\) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Here, \(n = 8\), \(a = 1\), and \(r = 2\):
\[ S_8 = 1 \cdot \frac{2^8 - 1}{2 - 1} = 2^8 - 1 = 256 - 1 = 255 \]

5. **Conclusion**:
Therefore, the value of \(n\) such that \((1+x)(1+x^2)(1+x^4)\cdots(1+x^{128}) = \Sigma_{r=0}^{n} x^r\) is \(n = 255\).

### Final Answer:
\[ n = 255 \]