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求数列n×2的n次方的前n项和sn
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人气:313 ℃ 时间:2020-01-29 15:55:45
解答
letS = 1.2 + 2.2^2+...+n.2^n(1)2S =1.2^2 + 2.2^2+...+n.2^(n+1)(2)(2)-(1)S = n.2^(n+1) - [2 + 2^2+...+2^n]S = n.2^(n+1) - 2(2^n-1)= 2+ (n-1).2^(n+1)Sn = S = 2+...
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