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已知数列{an}其通项公式为an=2的n次方分之2n-1 求数列的前n项和 Sn

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Sn=1/2^1+3/2^2+5/2^3+.+(2n-3)/2^(n-1)+(2n-1)/2^n ①‘

①×1/2 (1/2)Sn= 1/2^2+3/2^3+.+(2n-3)/2^n+(2n-1)/2^(n+1) ②

①-② (1/2)Sn=1/2+ 1/2 +1/2^2+.+1/2^(n-1)-(2n-1)/2^(n+1)

=1/2+ 1-1/2^(n-1)-(2n-1)/2^(n+1)

Sn =1 + 2 - 4/2^n-(2n-1)/2^n

∴ Sn=3-(2n+3)/2^n




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