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Question Number 47477 by naka3546 last updated on 10/Nov/18

$$Letn\in {\mathbb{Z}}^{+}$$$$f\left(1\right)=1$$$$f\left(2n\right)=f\left(n\right)$$$$f(2n+1)={\left(f\left(n\right)\right)}^2-2$$$$f\left(1\right)+f\left(2\right)+f\left(3\right)+\dots +f\left(100\right)=...$$

Answered by MJS last updated on 11/Nov/18

$$f\left(1\right)=1$$$$f\left(2\right)=f\left(2\times 1\right)=f\left(1\right)=1$$$$f\left(3\right)=f(2\times 1+1)=f{\left(1\right)}^2-2=-1$$$$f\left(4\right)=f\left(2\times 2\right)=f\left(2\right)=1$$$$f\left(5\right)=f(2\times 2+1)=f{\left(2\right)}^2-2=-1$$$$f\left(6\right)=f\left(2\times 3\right)=f\left(3\right)=-1$$$$f\left(7\right)=f(2\times 3+1)=f{\left(3\right)}^2-2=-1$$$$f\left(8\right)=f\left(2\times 4\right)=f\left(4\right)=1$$$$f\left(9\right)=f(2\times 4+1)=f{\left(4\right)}^2-2=-1$$$$f\left(10\right)=f\left(2\times 5\right)=f\left(5\right)=-1$$$$...$$$$\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{\forall }k\in \mathbb{N}\Rightarrow f\left(2^k\right)=1,\mathrm{for}\mathrm{all}\mathrm{other}$$$$\mathrm{values}f\left(n\right)=-1$$$$\Rightarrow f\left(1\right)+f\left(2\right)+f\left(4\right)+f\left(8\right)+f\left(16\right)+f\left(32\right)+f\left(64\right)=7$$$$7-93=-86\mathrm{is}\mathrm{the}\mathrm{answer}$$

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