`1+(3+1) (3^2+1) (3^4+1) (3^8+1) (3^16+1) (3^32+1)` is equal to

Options

A. `(3^64-1)/2`
B. `(3^64+1)/2`
C. `3^64-1`
D. `3^64+1`

Answer

To solve the expression \(1 + (3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)(3^{32} + 1)\), we can use the formula for the sum of a geometric series. Let's break it down step by step.

### Step 1: Rewrite the Expression
We start with the expression:
\[ 1 + (3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)(3^{32} + 1) \]
This can be rewritten as:
\[ 1 + 4(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)(3^{32} + 1) \]

### Step 2: Recognize the Pattern
Notice that each term can be expressed in the form \(3^{2^k} + 1\) where \(k\) ranges from 0 to 5. The expression can be simplified using the identity:
\[ a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \ldots + a + 1) \]
In our case, \(a = 3\) and \(n = 64\) (since \(2^6 = 64\)).

### Step 3: Apply the Identity
Using the identity, we can express:
\[ (3 - 1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)(3^{32} + 1) = 3^{64} - 1 \]
Thus,
\[ 4(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)(3^{32} + 1) = \frac{3^{64} - 1}{2} \]

### Step 4: Combine the Terms
Now, substituting back into our expression:
\[ 1 + 4(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)(3^{32} + 1) = 1 + \frac{3^{64} - 1}{2} \]
This simplifies to:
\[ 1 + \frac{3^{64}}{2} - \frac{1}{2} = \frac{2}{2} + \frac{3^{64}}{2} - \frac{1}{2} = \frac{3^{64} + 1}{2} \]

### Final Answer
Thus, the final result is:
\[ \frac{3^{64} + 1}{2} \]