1 + (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) is equal to =?

Answer: Option B

$$1+\left(3+1\right)$$  $$\left(3^2+1\right)$$ $$\left(3^4+1\right)$$ $$\left(3^8+1\right)$$ $$\left(3^{16}+1\right)$$ $$\left(3^{32}+1\right)$$
$$=1+\frac{1}{2}\left[\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\right]$$
$$=1+\frac{1}{2}\left[\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\right]$$
$$=1+\frac{1}{2}\left[\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\right]$$
$$\begin{array}{rl}& =1+\frac{1}{2}\left[\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\right]\\ & =1+\frac{1}{2}\left[\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\right]\\ & =1+\frac{1}{2}\left[\left(3^{32}-1\right)\left(3^{32}+1\right)\right]\\ & =1+\frac{1}{2}\left[\left(3^{64}+1\right)\right]\\ & =\frac{2+3^{64}-1}{2}\\ & =\frac{3^{64}+1}{2}\end{array}$$