神马作文网求和Sn=1/2+2/4+3/8+…+n/2^n,不明白Sn-1/2Sn是怎样减得
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求和Sn=1/2+2/4+3/8+…+n/2^n,不明白Sn-1/2Sn是怎样减得
不明白Sn-1/2Sn是怎样减得,
Sn-1/2Sn=1/2+1/4+1/8+....+1/2^n- n/2^(n+1) 怎么来的？
不明白Sn-1/2Sn是怎样减得,
Sn-1/2Sn=1/2+1/4+1/8+....+1/2^n- n/2^(n+1) 怎么来的？
Sn=1/2+2/4+3/8+…+n/2^n
1/2Sn= 1/4+2/8+…+(n-1)/2^n+n/2^(n+1)
Sn-1/2Sn=1/2+1/4+1/8+.+1/2^n- n/2^(n+1)
1/2Sn=1-1/2^n-n/2^(n+1)=[2^(n+1)-n-2]/2^(n+1)
Sn=[2^(n+1)-n-2]/2^n
=2-(n+2)/2^n
1/2Sn= 1/4+2/8+…+(n-1)/2^n+n/2^(n+1)
Sn-1/2Sn=1/2+1/4+1/8+.+1/2^n- n/2^(n+1)
1/2Sn=1-1/2^n-n/2^(n+1)=[2^(n+1)-n-2]/2^(n+1)
Sn=[2^(n+1)-n-2]/2^n
=2-(n+2)/2^n
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