Sn=(1+1)/2+(2+1)/2^2+(3+1)/2^3+``````+(n+1)/2^n=?
已知数列{an}的首项是a1=1,前n项和为Sn,且Sn+1=2Sn+3n+1(n∈N*).
已知:Sn=1+1/2+1/3+……+1/n,用数学归纳法证明:Sn^2>1+n/2(n>=2,n∈N+)
Sn=3+2^n Sn-1=3+2^(n-1).则Sn-Sn-1=?
数列Sn=(3n+1)/2-(n/2)an
已知数列{an}的首项a1=3,前n项和为Sn,且S(n+1)=3Sn+2n(n∈N)
已知an=(2n+1)*3^n,求Sn
an=(2^n-1)n,求Sn
已知Sn=2+5n+8n^2+…+(3n-1)n^n-1(n∈N*)求Sn
求和:Sn=1*n+2*(n-1)+3*(n-2)+……+n*1
数列an的前n项和Sn满足Sn=3n+1,n≤5,Sn=n^2,n≥6,求通项公式
Sn=1x2+3x2^2+5x2^3+…+(2n-1)x2^n sn=2sn-sn
(1).Sn=1+2×3+3×7...n(2^n-1),求Sn.