“an的通项公式为an+1”?
1.
a(n+1)=an+2
an=a1+2(n-1)=2n+1
2.
bn=an*3^n
= (2n+1)*3^n
Tn=3*3^1+5*3^2+7*3^3++(2n-3)*3^(n-2)+(2n-1)*3^(n-1)+(2n+1)*3^n
3Tn=3*3^2+5*3^3+7*3^4+……+(2n-3)*3^(n-1)+(2n-1)*3^n+(2n+1)*3^(n+1)
两式相减：
-2Tn=3*3^1+2*3^2+2*3^3+2*3^4+……+2*3^(n-1)+2*3^n-(2n+1)*3^(n+1)
=3+2*3^1+2*3^2+2*3^3+2*3^4+……+2*3^(n-1)+2*3^n-(2n+1)*3^(n+1)
=3+2*[3^n-1]/2-(2n+1)*3^(n+1)
=2+3^n-(2n+1)*3^(n+1)
=2-(6n+2)*3^n
-2Tn=2-(6n+2)*3^n
Tn=-1+(3n+1)*3^n