Câu hỏi của Phan Thuý An

Question

CMR :$\left(2+1\right).\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=2^{32}-1$

Nguyễn Huy Tú · Accepted Answer

$\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)$

$=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)$

$=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)$

$=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)$

$=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)$

$=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1$Câu trả lời: $\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)$

$=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)$

$=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)$

$=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)$

$=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)$

$=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1$

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