Question

已知数列{an},通项公式an=n(2^n-1),前n项和为Sn,则Sn+1/2n(n+1)=?

Answer 1

解：因为 an=n(2^n-1)=n*2^n-n
所以 Sn=1*2+2*2^2+3*2^3+..............+n*2^n -(1+2+...+n) (1)
2Sn= 1*2^2+2^2^3+........+(n-1)*2^n+n*2^(n+1)-2(1+2+3+...+n) (2)
(2)-(1)：sn=-2-2^2-......-2^n+n*2^(n+1)-(1+2+3+...+n)
=2(1-2^n)+n*2^(n+1)-n(n+1)/2
=(n-1)*2^(n+1)-1/2*n(n+1)+2
因此 Sn+1/2n(n+1)=(n-1)*2^(n+1)+2

Answer 2

由题知：an=n(2^n-1)
所以an/n=(2^n-1)
即数列{an/n}为等比数列，公比q=2，首项a1=1
所以数列{an/n}的前N+1项和（Sn+1/n+1）=2^(n+1）-1
所以Sn+1/2n(n+1)=[2^（n+1）-1]/2n

Answer 3

∵an=n(2^n-1)，∴a1=1,a2=2*2,a3=3*2^2……
∴Sn=1+2*2+3*2^2+4*2^3+……+n*(2^n-1)
∴2Sn=1*2+2*2^2+3*2^3+……+(n-1)*2^n-1+n*2^n
Sn-2Sn=1+2+2^2+2^3+……+2^(n-1)-n*2^n=((1+2n)/3 )-n*2^n=-Sn
∴Sn=n*2^n-(1+2n)/3
∴Sn+1=((n+1)*2^n+1)-(1+2^n+1/3
∴所求=(2^n/n) -（1+2^n+1）/6n(n+1)

Answer 4

an=n(2^n-1)=n*2^n-n=2n*2^(n-1)-n=(2n-2)*2^(n-1)-n+2^n=2(n-1)*2^(n-1)+(-2n+2)+n-2+2^n=2*an-1 +n+2^n-2
..
a2=2a1+2+4-2
a1=a1+1+2-2-1
Sn=2Sn-1+a1+(1+2+..+n)+(2+4+..+2^n)-2n-1
a1=1
Sn=2Sn-1+(1+2+..+n)+(2+4+..+2^n)-2n=2Sn-1+(1+n)n/2+2*(2^n-1)-2n
Sn=an+Sn-1
an+Sn-1=2Sn-1 +(n+1)n/2+2*(2^n-1)-2n
Sn-1=an-(n+1)n/2-2*(2^n-1)+2n
an=n*(2^n-1)
Sn-1= -(n+1)n/2+(n-2)*(2^n-1)+2n
Sn+1= -(n+3)(n+2)/2+n*[2^(n+2)-1]+2n+4
Sn+1/2n(n+1)= -(n+3)(n+2)/[4n(n+1)] +n*[2^(n+2)-1]/[2n(n+1)]+(2n+4)/[2n(n+1)]