Sn=1+(1/1+2)+(1/1+2+3)+(1/1+2+3+4 )+...+(1/1+2+...+n)=
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=?
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+)
设Sn=-1+3-5+7-…+(-1)n(2n-1),则Sn=______.
已知数列{an}的首项a1=3,前n项和为Sn,且S(n+1)=3Sn+2n(n∈N)
a1=1,Sn为an前n项和,Sn-S(n-1)=√Sn-√S(n-1)(n>=2)
设Sn是数列an的前n项和,已知a1=1,an=-Sn*Sn-1,(n大于等于2),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
已知数列{an}的前n项和为Sn,a1=-23,Sn+1Sn=an-2(n≥2,n∈N)