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

1^{(3⁶⁴ - 1)}/₂
2^{(3⁶⁴ + 1)}/₂
33⁶⁴ - 1
43⁶⁴ + 1

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 1 + (3 + 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1) \\ = 1 + \frac{1}{2} [(3 - 1) (3 + 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{2} - 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{4} - 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{8} - 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{16} - 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{32} - 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{64} + 1)] \\ = \frac{2 + 3^{64} - 1}{2} \\ = \frac{3^{64} + 1}{2} \end{aligned}

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