求2N-1/2^N的前N项和

来源：百度知道 编辑：UC知道 时间：2024/06/04 15:35:08
同上!!!!!!!!!!在线等速度,带过程!!!!!!!!!!!
是(2N-1)除(2^N)啊!!!

S=2^(-N) * (-3 + 3 2^N - 2 N)

S=Sum[k=1..n,(2k-1)/2^k]
=Sum[k=1..n,(2k-1)2^(n-k)/(2^n)]
=Sum[k=1..n,(2k-1)2^(n-k)]/(2^n)
=Sum[k=1..n,2^(n-k+1)k - 2^(n-k)]/(2^n)
=-Sum[k=1..n,2^(k-1)]/(2^n) +
Sum[k=1..n,2^(n-k+1)k]/(2^n)
=-(-1 + 2^n)/2^n +2^(1 - n) (-2 + 2^(1 + n) - n)/2^n
=2^(-N) * (-3 + 3 2^N - 2 N)

2N-1/2^N
当作两个数列来计算
Sn=(2+4+6+...+2n)-(1/2+1/4+1/6+...+1/2^n)
=n(n+1)-[1-(1/2)^n]
=n^2+n-1+1/2^n

Sn=(2+4+6+……+2n)+(1/2+1/4+……+1/2^n)
=n(2+2n)/2+[1/2-1/2^(n+1)]/(1-1/2)
=n(n+1)+(1/2^n-1)
=n^2+n+2^(-n)-1

分成两部分,

2n前n项和,即前n个偶数和= (2+2n)*(n/2),

1/2^N 前n项和 ,是等比数列,=(1/2)*(1-(1/2)n)/(1-1/2)=1-2^(-n),

故 2N-1/2^N的前N项和=(2+2n)*(n/2)-1+2^(-n)

2N-1/2^N的前N项和
=2(1+2+3+....+N)-(1/2+1/4+....1/2^N)
=N(N+1)-1+1/2^N

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