求数列-1,2,-3,4,-5…的前n项和

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求数列-1,2,-3,4,-5…的前n项和求数列-1,2,-3,4,-5…的前n项和求数列-1,2,-3,4,-5…的前n项和an=(2n-1)/2^nSn=1/2+3/4+5/8+...+(2n-3

求数列-1,2,-3,4,-5…的前n项和
求数列-1,2,-3,4,-5…的前n项和

求数列-1,2,-3,4,-5…的前n项和

an=(2n-1)/2^n
Sn=1/2+3/4+5/8+...+(2n-3)/2^(n-1)+(2n-1)/2^n
1/2Sn= 1/4+3/8+... +(2n-3)/2^n+(2n-1)/2^(n+1)
上式减下式:
Sn-1/2Sn=1/2+2/4+2/8+2/16+... + 2/2^n-(2n...

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an=(2n-1)/2^n
Sn=1/2+3/4+5/8+...+(2n-3)/2^(n-1)+(2n-1)/2^n
1/2Sn= 1/4+3/8+... +(2n-3)/2^n+(2n-1)/2^(n+1)
上式减下式:
Sn-1/2Sn=1/2+2/4+2/8+2/16+... + 2/2^n-(2n-1)/2^(n+1)
=1/2-(2n-1)/2^(n+1)-1+2(1/2^1+1/2^2+1/2^3+...+1/2^n)
=1/2-(2n-1)/2^(n+1)-1+(1-1/2^n)/(1-1/2)
=-1/2-(2n-1)/2^(n+1)+2-1/2^(n-1)
=3/2-[(2n-1)/4+1]/2^(n-1)
=3/2-(2n+3)/2^(n+1)
于是1/2Sn=3/2-(2n+3)/2^(n+1)
Sn=3-(2n+3)/2^n
求采纳为满意回答。

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