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Question Number 24286 by Joel577 last updated on 15/Nov/17

$$\mathrm{If}\mid x\mid <1\mathrm{then}$$ $$(x+1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1).....$$ $$\mathrm{is}\mathrm{equal}\mathrm{to}$$

Answered by mrW1 last updated on 15/Nov/17

$$let{P}_n=(x+1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1).....(x^{2^n}+1)$$ $$(x-1)P_n=(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1).....(x^{2^n}+1)$$ $$(x-1)P_n=(x^2-1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1).....(x^{2^n}+1)$$ $$(x-1)P_n=(x^4-1)(x^4+1)(x^8+1)(x^{16}+1).....(x^{2^n}+1)$$ $$......$$ $$(x-1)P_n=(x^{2^n}-1)(x^{2^n}+1)$$ $$(x-1)P_n=x^{2^{n+1}}-1$$ $$\Rightarrow P_n=\frac{1-x^{2^{n+1}}}{1-x}$$ $$\underset{n\to \mathrm{\infty }}{\lim }{P}_n=\frac{1}{1-x}$$ $$\Rightarrow (x+1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1).....=\frac{1}{1-x}$$

Commented bymath solver last updated on 16/Nov/17

$$\mathrm{sir},\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{your}\mathrm{last}\mathrm{step}$$ $$\mathrm{when}\mathrm{n}\mathrm{tending}\mathrm{to}\mathrm{infinity}?$$

Commented bymrW1 last updated on 16/Nov/17

$$wehave{P}_n=\frac{1-x^{2^{n+1}}}{1-x}$$ $$whenn\to \mathrm{\infty },$$ $$2^{n+1}\to \mathrm{\infty }$$ $$since\mid x\mid <1$$ $$x^{2^{n+1}}\to 0$$ $$P_n\to \frac{1-0}{1-x}=\frac{1}{1-x}$$

Commented bymath solver last updated on 16/Nov/17

$$\mathrm{okk},\mathrm{i}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{read}\mid \mathrm{x}\mid <1.$$

Commented byJoel577 last updated on 16/Nov/17

$$thankyouverymuch$$

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