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$$1+(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)$$

Ángel Mario Gallegos
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mkumar
  • 9
  • If one-line answers are rarely good, so too with one-line questions. It would greatly improve your post to indicate what you've tried, or even what makes such an exercise interesting to you. – hardmath Apr 24 '16 at 15:09

2 Answers2

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We have \begin{align} &1+(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\ &\quad=1+\frac{(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)}{2}\\ &\quad=1+\frac{3^{64}-1}{2}\\ &\quad=\frac{3^{64}+1}{2}. \end{align}

Solumilkyu
  • 3,289
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Let $$S=1+(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)$$ Then, \begin{align*} (3-1)S&=(3-1)+(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\[3pt] 2S&=2+(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\[3pt] &=2+(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\\[3pt] &=2+(3^{16}-1)(3^{16}+1)(3^{32}+1)\\[3pt] \vdots&=\vdots\\ &=2+3^{64}-1\\ &=3^{64}+1 \end{align*} Hence $$\boxed{\color{blue}{S=\frac{3^{64}+1}{2}}}$$

Ángel Mario Gallegos
  • 19,882