计算sn=1+1/(1+2)+1/(1+2+3)+...1/(1+2+3+...+n)=
(1).Sn=1+2×3+3×7...n(2^n-1),求Sn.
Sn=1x2+3x2^2+5x2^3+…+(2n-1)x2^n sn=2sn-sn
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*).
Sn=3+2^n Sn-1=3+2^(n-1).则Sn-Sn-1=?
计算sn=1*3+3*5+5*7+...+(2n-1)(2n+1)
已知数列【An】的前n项和为Sn,A1=-3分之2,满足Sn+Sn分之1+2=An(n大于等于2).计算S1,S2,S3
an=(2^n-1)n,求Sn
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+)
已知数列{an}的首项a1=3,前n项和为Sn,且S(n+1)=3Sn+2n(n∈N)
设Sn=-1+3-5+7-…+(-1)n(2n-1),则Sn=______.
已知a1=3,an=Sn-1+2^n(n大于等于2),求an,Sn?
已知数列{an}的首相a1=1,a2=3,前n项和为Sn,且Sn+1(下标)、Sn、Sn-1(下标)(n≥2)满足(Sn