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运用平方差公式计算(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)
^是乘方
人气:178 ℃ 时间:2019-08-20 01:51:05
解答
(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)
=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)/2
=(3^4-1)(3^4+1)(3^8+1)...(3^2n+1)/2
……
=(3^4n-1)/2
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