Question

设Sn是数列{an}的前n项和，已知a1=3，an+1=2Sn+3，求{an}的通项公式和令bn=(2n-1

设Sn是数列{an}的前n项和，已知a1=3，an+1=2Sn+3，求{an}的通项公式和令bn=(2n-1)an，求数列{bn}的前n项和Tn

Answer 1

a(n+1)= 2Sn+3
for n>=2
an = Sn - S(n-1)
an = (1/2)( a(n+1)-3 ) - (1/2)( an -3 )
2an = a(n+1) - an
a(n+1) = 3an
an= 3^(n-1) .a1
=3^n

let
S= 1.3^1 +2.3^2+....+n.3^n (1)
3S= 1.3^2 +2.3^3+....+n.3^(n+1) (2)
(2)-(1)
2S =n.3^(n+1) -( 3^1+3^2+...+3^n)
=n.3^(n+1) - (3/2)( 3^n-1)

bn=(2n-1)an
=(2n-1).3^n
=2(n.3^n) - 3^n
Tn =b1+b2+...+bn
=2S - (3/2)(3^n-1)
=n.3^(n+1) - (3/2)( 3^n-1) -(3/2)(3^n-1)
=n.3^(n+1) - 3( 3^n-1)
= 3 + (3n-3).3^n