已知数列｛an｝满足a1=2,a n+1=2an+3. （2）求数列｛nan｝的前n项和，为么是a1+3⼀5

a(n+1)=2an+3
a(n+1)+3=2an+6=2(an+3)
所以数列｛an+3｝是公比为2的等比数列
首项为a1+3=5
则an+3=(a1+3)*2^(n-1)=5*2^(n-1)
所以an=5*2^(n-1)-3

需要求Sn吗？
Sn=(5*2^0-3)+(5*2^1-3)+(5*2^2-3)+……+[5*2^(n-1)-3]
=5*[2^0+2^1+2^2+……+2^(n-1)]-3n
=5*1*(1-2^n)/(1-2)-3n
=5*2^n-3n-5

设Tn=1*2^0+2*2^1+3*2^2+……+n*2^(n-1)
2Tn=1*2^1+2*2^2+3*2^3+……+n*2^n
-Tn=Tn-2Tn
=2^0+2^1+2^2+……+2^(n-1)-n*2^n
=1*(1-2^n)/(1-2)-n*2^n
=(1-n)*2^n-1
所以Tn=(n-1)*2^n+1
则1*a1+2*a2+……+n*an
=1*(5*2^0-3)+2*(5*2^1-3)+……+n*[5*2^(n-1)-3]
=5*[1*2^0+2*2^1+……+n*2^(n-1)]-3*(1+2+……+n)
=5*Tn-3n(n+1)/2
=5(n-1)*2^n+5-3n(n+1)/2

由a(n+1)=2an+3有a(n+1)+3=2an+6=2(an+3)，即[a(n+1)+3]/(an+3)=2
令bn=an+3，数列{bn}为公比q=2的等比数列，易知其首项b1=5
则bn=b1*q^(n-1)=5*2^(n-1)，数列{bn}前n项Sn(bn)=5*(1-2^n)/(1-2)=5*2^n-5
即有an=bn+3=5*2^(n-1)+3
所以对于数列{an}，前n项和Sn(an)=Sn(bn)+3n=5*2^n+3n-5