If |x|

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(B) Let $P_n = (1 + x)(1 + x^2)(1 + x^4) \dots (1 + x^{2^n})$.Multiply and divide by $(1 - x)$:$P_n = \frac{(1 - x)(1 + x)(1 + x^2)(1 + x^4) \dots (1

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$\frac{1}{1 - x}$

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(B) Let $P_n = (1 + x)(1 + x^2)(1 + x^4) \dots (1 + x^{2^n})$.Multiply and divide by $(1 - x)$:$P_n = \frac{(1 - x)(1 + x)(1 + x^2)(1 + x^4) \dots (1 + x^{2^n})}{1 - x}$Using the identity $(1 - a)(1 + a) = (1 - a^2)$,we get:$P_n = \frac{(1 - x^2)(1 + x^2)(1 + x^4) \dots (1 + x^{2^n})}{1 - x} = \frac{(1 - x^4)(1 + x^4) \dots (1 + x^{2^n})}{1 - x}$Continuing this process,we obtain:$P_n = \frac{1 - x^{2^{n+1}}}{1 - x}$Since $|x| Therefore,$\lim_{n 	o \infty} P_n = \frac{1 - 0}{1 - x} = \frac{1}{1 - x}$.

If $|x| < 1$,then $\lim_{n \to \infty} \{(1 + x)(1 + x^2)(1 + x^4) \dots (1 + x^{2^n})\}$ is equal to

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