已知数列｛an｝满足a1=2,an+1+2an+3 1）求｛an｝的通项公式； 2）求数列｛Nan｝的前n项和Sn；

来源：学生作业学帮网编辑：学帮网时间：2024/04/26 01:55:45

原题有误,我猜测应该是a(n+1)=2an+3

第1问：
a(n+1)=2an+3
a(n+1)+3=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=5*2^(n-1)
所以an=5*2^(n-1)-3

第2问：
令Tn=1*(a1+3)/5+2*(a2+3)/5+3*(a3+3)/5+……+n*(an+3)/5
=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=2Tn-Tn
=-1*2^0-1*2^1-1*2^2-……-1*2^(n-1)+n*2^n
=-1*(1-2^n)/(1-2)+n*2^n
=(n-1)*2^n+1
Sn=1*a1+2*a2+3*a3+……+n*an
=1*(5*2^0-3)+2*(5*2^1-3)+3*(5*2^2-3)+……+n*[5*2^(n-1)-3]
=5*[1*2^0+2*2^1+3*2^2+……+n*2^(n-1)]-3*(1+2+3+……+n)
=5*Tn-3*n*(n+1)/2
=5(n-1)*2^n+5-3n(n+1)/2

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