Question

设数列〔an〕满足a1=1，a2=5/3(3分之5），an+2=5/3an+1－2/3an,(n属于N※）. 求数列{nan}的前n项和Sn

（算出AN后求SN的具体啊）

Answer 1

3a(n+2)=5a(n+1)-2an3a(n+2)-3a(n+1)=2a(n+1)-2an[a(n+2)-a(n+1)]/[a(n+1)-an]=2/3a(n+1)-an=(a2-a1)(2/3)^(n-1)=(2/3)^na2-a1=2/3a3-a2=(2/3)^2a4-a3=(2/3)^3……a(n-1)-a(n-2)=(2/3)^(n-2)an-a(n-1)=(2/3)^(n-1)将上列式子相加得an-a1=2/3+(2/3)^2+(2/3)^3+……(2/3)^(n-1)=(2/3)[1-(2/3)^(n-1)]/(1-2/3)an=3-2(2/3)^(n-1)nan=3n-2n(2/3)^(n-1)Sn=3(1+2+3+……+n)-2[1+2*(2/3)+3*(2/3)^2+……+(n-1)(2/3)^(n-2)+n(2/3)^(n-1)]=3n(n+1)/2-2[1+2*(2/3)+3*(2/3)^2+……+(n-1)(2/3)^(n-2)+n(2/3)^(n-1)](2/3)Sn=n(n+1)-2[(2/3)+2*(2/3)^2+……+(n-2)(2/3)^(n-2)+(n-1)(2/3)^(n-1)+n(2/3)^n]Sn-(2/3)Sn=Sn/3=n(n+1)/2-2[1-n(2/3)^n+(2/3)+(2/3)^2+……+(2/3)^(n-2)+(2/3)^(n-1)]=n(n+1)/2-2[1-n(2/3)^n+(2/3)(1-(2/3)^(n-1))/(1-2/3)]=n(n+1)/2-2[1-n(2/3)^n+2-(2/3)^n]=n(n+1)/2+2(n+1)(2/3)^n-6Sn=3n(n+1)/2+6(n+1)(2/3)^n-18