$3(2^2+1)(2^4+1)…(2^{64}+1)+1$ $= (2^2-1)(2^2+1)…(2^{64}+1)+1$ $= (2^4-1)(2^4+1)…(2^{64}+1)+1$ $= (2^8-1)(2^8+1)…(2^{64}+1)+1$ $= (2^{64}-1)(2^{64}+1)+1$ $= 2^{128}-1+1$ $= 2^{128}$ Trả lời
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