Question-203634

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

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Answered by Frix last updated on 24/Jan/24

((2^n^2 −1)/(2^n −1))≈2^(n^2 −n) ⇒ k=271

$$\frac{\mathrm{2}^{{n}^{\mathrm{2}} } −\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}}\approx\mathrm{2}^{{n}^{\mathrm{2}} −{n}} \:\Rightarrow\:{k}=\mathrm{271} \\ $$

Commented by Frix last updated on 24/Jan/24

Sorry I was confused. N=((2^(17^2 ) −1)/(2^(17) −1))≈2^(271) But because N=2k+1 ⇒ a_1 =0 ⇒ k=272

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

Answered by witcher3 last updated on 24/Jan/24

17=(17)^2 x^(2n+1) +1=(x+1)(Σ_(k=0) ^(2n) (−1)^k x^k ) (((2^(17) )^(17) +1)/(2^(17) +1))=Σ_(k=0) ^(16) (−1)^k 2^(17k) a_k =17.16=272

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

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Question-203634

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