数列求和习题：Sn＝1/2+3/4+5/8+……+2n-1/2的n次方 求Sn
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数列求和习题：Sn＝1/2+3/4+5/8+……+2n-1/2的n次方 求Sn
数列求和习题：Sn＝1/2+3/4+5/8+……+2n-1/2的n次方 求Sn
数列求和习题：Sn＝1/2+3/4+5/8+……+2n-1/2的n次方 求Sn
解析：∵Sn＝1/2+3/4+5/8+……+（2n-1）/2^n
2 Sn＝1+3/2+5/4+7/8+……+(2n-1)/2^(n-1)
∴Sn=1+2/2+2/4+2/8+……+2/2^(n-1)-(2n-1)/2^n
=1+1+1/2+1/4+1/8+……+1/2^(n-2)-(2n-1)/2^n
=1+[1-(1/2)^(n-1)]/(1-1/2)-(2n-1)/2^n
=1+2-1/2^(n-2))-(2n-1)/2^n
=3-(2n+3)/2^n
∵Sn＝1/2+3/4+5/8+……+(2n-1)/2^n.......(1)
∴Sn/2=1/4+3/8+5/16+……+(2n-3)/2^n+(2n-1)/2^(n+1).......(2)
==>Sn/2=Sn-Sn/2
=1/2+2/4+2/8+2/16+……+2/2^n-(2n-1)/2^(n+1)
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∵Sn＝1/2+3/4+5/8+……+(2n-1)/2^n.......(1)
∴Sn/2=1/4+3/8+5/16+……+(2n-3)/2^n+(2n-1)/2^(n+1).......(2)
==>Sn/2=Sn-Sn/2
=1/2+2/4+2/8+2/16+……+2/2^n-(2n-1)/2^(n+1)
=1/2+1/2+1/4+1/8+……+1/2^(n-1)-(2n-1)/2^(n+1)
=-1/2+[1+1/2+1/4+1/8+……+1/2^(n-1)]-(2n-1)/2^(n+1)
=-1/2+(1-1/2^n)/(1-1/2)-(2n-1)/2^(n+1)
=-1/2+2-1/2^(n-1)-(2n-1)/2^(n+1)
=3/2-(2n+3)/2^(n+1)
故Sn=2[3/2-(2n+3)/2^(n+1)]
=3-(2n+3)/2^n
收起
sn={2^(n-1)+3*2^(n-2)+5*2^(2n-3)....2n-1}/2n
2sn={2^n+3*2^(n-1)+....+2*(2n-1)}/2n
下面减去上面的式子
得 s={2^n+2*2^n-1)+2*2^(n-2)....2*2^0-2n+1}/2n
=2-(2n+1)/2^n+{1+1/2+1/4......1/ 2^(2n-1)...
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sn={2^(n-1)+3*2^(n-2)+5*2^(2n-3)....2n-1}/2n
2sn={2^n+3*2^(n-1)+....+2*(2n-1)}/2n
下面减去上面的式子
得 s={2^n+2*2^n-1)+2*2^(n-2)....2*2^0-2n+1}/2n
=2-(2n+1)/2^n+{1+1/2+1/4......1/ 2^(2n-1)}
=3-(2n+1)/ 2^n
这太难书写了，还有高中的都快忘没了，对了你得加分！！！！！！！！！
收起
Sn=1/2+3/4+5/8+……+2n-1/2^n,则
1/2*Sn=1/4+3/8+5/16++……+2n-1/2^(n+1)
错位相减，得
1/2*Sn=1/2+2*(1/4+1/8+……+1/2^n)-(2n-1)/2^(n+1)
Sn=3-(2n+3)/2^n