Question

数列题一道~

a(n)=3a(n-1)+1
b(n)=a(n)+0.5
a(1)=1
去演b(n)的前n项和

Answer 1

a(n)=3a(n-1)+1
a(n)+1/2=3a(n-1)+1+1/2=3[a(n-1)+1/2)]
[a(n)+1/2]/[a(n-1)+1/2)]=3
数列{a(n)+1/2}为首项3/2,公比为3的等比数列
a(n)+1/2=(3/2)*3^(n-1)=(3^n)/2
a(n)=(3^n)/2-1/2
b(n)=a(n)+0.5=(3^n)/2
b(n)的前n项和=3(3^n-1)/4

Answer 2

an=3a(n-1)+1
an+0.5=3(a(n-1)+0.5)
=3^2(a(n-2)+0.5)
=……
=3^(n-1)(a1+0.5)
=3^(n-1)*1.5
=0.5*3^n

an=0.5*3^n-0.5

bn=an+0.5=0.5*3^n

∑bn=3(3^n-1)/2