Question

高二数学，求解，，，，！

Answer 1

a1=1 ,an=1/3*a(n-1)+(1/3)^n
an=1/3*a(n-1)+(1/3)^n
等式两边除以(1/3)^n得
an/(1/3)^n=a(n-1)/(1/3)^n+1
[an/(1/3)^n]-[]a(n-1)/(1/3)^(n-1)]=1
{an/(1/3)^n}是等差数列
an/(1/3)^n=a1/(1/3)+(n-1)=1/(1/3)+(n-1)=n+2
an/(1/3)^n=n+2
an=(n+2)*(1/3)^n
2)Sn=a1+a2+...+an
Sn=1+4*(1/3)^2+5*(1/3)^3+6*(1/3)^4...+(n+2)*(1/3)^n
1/3*Sn=1/3+4*(1/3)^3+5*(1/3)^4+...+(n+1)*(1/3)^n+(n+2)*(1/3)^(n+1)
两式相减得

2/3*Sn=2/3+4*(1/3)^2+(1/3)^3+(1/3)^4+(1/3)^n-(n+2)*(1/3)^(n+1)
2/3*Sn=2/3+[1/3+(1/3)^2+(1/3)^3+(1/3)^4+(1/3)^n]-(n+2)*(1/3)^(n+1)
2/3*Sn=2/3+[1/3(1-(1/3)^n]/(1-1/3)-(n+2)*(1/3)^(n+1)
2/3*Sn=2/3+[1/2(1-(1/3)^n]-(n+2)*(1/3)^(n+1)
2/3*Sn=7/6-[(2n+7)/2](1/3)^n
Sn=7/4-[(6n+21)/4]*(1/3)^n