Question

已知数列{an}满足a1=2,an+1=2an+3.求数列{nan}的前n项和

Answer 1

a(n+1)=2an+3
∴a(n+1)+3=2an+6
∴a(n+1)+3=2(an+3)
∴﹛an+3﹜是等比数列
∴an+3=(a1+3)×2^(n-1)=5×2^(n-1)
∴an=5×2^(n-1)-3
∴nan=5n×2^(n-1)-3n
∴Sn=5×[1×2^0+2×2^1+...+n×2^(n-1)]-3(1+2+3+...+n)
=5×[1×2^0+2×2^1+...+n×2^(n-1)]-3×n(n+1)/2
设Tn=1×2^0+2×2^1+...+n×2^(n-1)
∴2Tn=1×2^1+2×2^2+..+(n-1)×2^(n-1)+n×2^n
两式相减得：-Tn=2^0+2^1+2^2+..+2^(n-1)-n×2^n
=2^0×(1-2^n)/(1-2)-n×2^n
=2^n-1-n×2^n
=(1-n)×2^n-1
∴Tn=(n-1)×2^n+1
∴Sn=5(n-1)×2^n-5-3n(n+1)/2