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Question Number 218673 by Spillover last updated on 14/Apr/25

Commented by Nicholas666 last updated on 14/Apr/25

$$\phi$$

Answered by Nicholas666 last updated on 14/Apr/25

$$\frac{\phi }{{\phi }^2-1}+\frac{{\phi }^2}{{\phi }^4-1}+\frac{{\phi }^4}{{\phi }^8-1}+.....$$$$solution;$$$$\frac{{\phi }^{2n-1}}{{\phi }^{2n}-1}=\frac{1}{{\phi }^{2n}-1}-\frac{1}{{\phi }^{2n}-1}$$$$S=(\frac{1}{\phi -1}-\frac{1}{\phi -1})+(\frac{1}{{\phi }^2-1}-\frac{1}{{\phi }^4-1})+(\frac{1}{{\phi }^4-1}-\frac{1}{{\phi }^8-1})+......$$$$S_n=\frac{1}{\phi -1}-\frac{1}{{\phi }^{2n}-1}$$$$S_n=\phi -\frac{1}{{\phi }^{2n}-1}=\phi$$

Answered by Charleston last updated on 14/Apr/25

Answered by Spillover last updated on 14/Apr/25

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