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等比数列(n×2*n),q=2,求前n项和
人气:352 ℃ 时间:2020-05-14 02:22:03
解答
let
S = 1.2^1+2.2^2+.+n.2^n (1)
2S = 1.2^2+2.2^3+.+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) - (2+2^2+...+2^n)
=n.2^(n+1) - 2(2^n -1)
= 2+ (2n-2).2^n
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