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iceghost

$(x+1)(x^2+1)(x^4+1)\cdots(x^{32}+1) \\
=\dfrac{(x-1)(x+1)(x^2+1)(x^4+1)...(x^{32}+1)}{x-1} \\
=\dfrac{(x^2-1)(x^2+1)(x^4+1)...(x^{32}+1)}{x-1} \\
\qquad \qquad \vdots \\
=\dfrac{(x^{32}-1)(x^{32}+1)}{x-1} \\
=\dfrac{x^{64}-1}{x-1}$
 
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