What is the simplified value of `(3+1)(3^(2)+1)(3^(4)+1)(3^(8)+1)(3^(16)+1)` ?

Options

A. `((3^(32)-1))/(2)`
B. `((3^(16)-1))/(2)`
C. `((3^(64)-1))/(2)`
D. `((3^(128)-1))/(2)`

Answer

To simplify the expression \((3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\), we can use the identity that relates products of terms of the form \(a^n + 1\).

### Step-by-Step Solution:

1. **Write down the expression:**
\[ (3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1) \]

2. **Recognize the pattern:**
Notice that each term can be expressed as \(3^{2^k} + 1\) where \(k\) ranges from 0 to 4.

3. **Use the identity:**
We can use the identity:
\[ (a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) = \frac{a^{32} - 1}{a - 1} \]
for \(a = 3\).

4. **Apply the identity:**
Substitute \(a = 3\) into the identity:
\[ (3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1) = \frac{3^{32} - 1}{3 - 1} \]

5. **Simplify the expression:**
Since \(3 - 1 = 2\), we can simplify further:
\[ = \frac{3^{32} - 1}{2} \]

6. **Final result:**
Thus, the simplified value of the expression is:
\[ \frac{3^{32} - 1}{2} \]