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Question Number 203634 by cortano12 last updated on 24/Jan/24

Answered by Frix last updated on 24/Jan/24

$$\frac{2^{n^2}-1}{2^n-1}\approx 2^{n^2-n}\Rightarrow k=271$$

Commented by Frix last updated on 24/Jan/24

$$\mathrm{Sorry}\mathrm{I}\mathrm{was}\mathrm{confused}.$$$$N=\frac{2^{{17}^2}-1}{2^{17}-1}\approx 2^{271}$$$$\mathrm{But}\mathrm{because}N=2k+1\Rightarrow a_1=0\Rightarrow k=272$$

Answered by witcher3 last updated on 24/Jan/24

$$17={\left(17\right)}^2$$$${\mathrm{x}}^{2\mathrm{n}+1}+1=(\mathrm{x}+1)\left(\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\right)$$$$\frac{{\left(2^{17}\right)}^{17}+1}{2^{17}+1}=\underset{\mathrm{k}=0}{\sum ^{16}}{(-1)}^{\mathrm{k}}{2}^{17\mathrm{k}}$$$${\mathrm{a}}_{\mathrm{k}}=17.16=272$$

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