Question

已知数列{an}中,an=n(2的n次方-1),其前n项和为Sn,则Sn+1/2n(n+1)等于?

Answer 1

an=n(2^n-1)
an=n*2^n-n
a1=1*2^1-1
a2=2*2^2-2
a3=3*3^3-3
.
an=n*2^n-n
Sn=a1+a2+a3+.+an
=1*2^1-1+2*2^2-2+3*3^3-3+.+n*2^n-n
=1*2^1+2*2^2+3*3^3+.+n*2^n-(1+2+3+...+n)
=1*2^1+2*2^2+3*3^3+.+n*2^n-(1+n)*n/2
2Sn=1*2^2+2*2^3+3*3^4+.+(n-1)*2^n+n*2^(n+1)-(1+n)*n
Sn-2Sn=2^1+2^2+2^3+.+2^n-n*2^(n+1)+(1+n)*n/2
=2(1-2^n)/(1-2)-n*2^(n+1)+(1+n)*n/2
=2^(n+1)-2-n*2^(n+1)+(1+n)*n/2
=2^(n+1)(1-n)-2+(1+n)*n/2
Sn=2^(n+1)(n-1)+2-(1+n)*n/2
Sn+1/2n(n+1)
=2^(n+1)*(n-1)+2-(1+n)*n/2+1/2*n(n+1)
=(n-1)*2^(n+1)+2