计算(7+1)(7^2+1)(7^4+1)(7^8+1)(7^16+1)
2个回答
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(7 + 1)(7^2 + 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7 - 1)(7 + 1)(7^2 + 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^2 - 1)(7^2 + 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^4 - 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^8 - 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^16 - 1)(7^16 + 1)
= (1/6) × (7^32 - 1)
= (7^32 - 1)/6
= (1/6) × (7 - 1)(7 + 1)(7^2 + 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^2 - 1)(7^2 + 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^4 - 1)(7^4 + 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^8 - 1)(7^8 + 1)(7^16 + 1)
= (1/6) × (7^16 - 1)(7^16 + 1)
= (1/6) × (7^32 - 1)
= (7^32 - 1)/6
追问
1/6是哪来滴?
追答
因为前面乘了(7 - 1) ,也就是6,为了使等式的值不变,所以还要乘以 (1/6)
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