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Question Number 95604 by 174 last updated on 26/May/20

Commented by MJS last updated on 26/May/20

$$x^5+1=(x+1)(x^2-\frac{1+\sqrt{5}}{2}x+1)(x^2-\frac{1-\sqrt{5}}{2}x+1)$$$$\mathrm{decompose}\mathrm{and}\mathrm{solve}.$$

Commented by peter frank last updated on 26/May/20

$$\mathrm{sir}\mathrm{can}\mathrm{show}\mathrm{how}\mathrm{to}\mathrm{partialize}$$$$\mathrm{given}\mathrm{equation}\mathrm{of}$$$$\mathrm{decomposition}.$$

Commented by Tony Lin last updated on 26/May/20

$${(-1)}^5+1=0$$$$\to x^5+1=(x+1)(x^4-x^3+x^2-x+1)$$$$let{x}^4-x^3+x^2-x+1$$$$=(x^2+ax+1)(x^2+bx+1)$$$$andsolveaandb$$

Answered by mathmax by abdo last updated on 27/May/20

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}\mathrm{I}={\int }_0^1\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{5\mathrm{n}}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{5\mathrm{n}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{5\mathrm{n}+1}$$$$=1-\frac{1}{6}+\frac{1}{11}-\frac{1}{16}+\frac{1}{21}-....$$

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