Question

已知{an}为等比数列,其前n项和为Sn,且Sn=2∧n+a,若bn=(2n-1)an,求bn的前

Answer 1

1,求出an通项公式

an = Sn - Sn-1 = (2^n+a) - (2^n-1 +a) = 2^(n-1)

q = an/an-1 = 2^(n-1)/2^(n-2) = 2

由等比数列求和公式

Sn = (a1- an*q)/(1-q) = 2^n -a1

所以 -a1 = a

a1 = S1 = 2^1 + a = 2+a

所以 2+a = -(-a1) = -a

a = -1

a1 = 1

所以 an = 1* 2^(n-1) = 2^(n-1)

2

设{Cn} = {2n-1},Cn公差 d = 2

q Tn - Tn = (c1a2+c2a3+c3a4+...+ cnan+1) - (c1a1+c2a2+c3a3+...+ cnan)

= -c1a1 + (-d)a2 + (-d)a3+...+ (-d) an + cnan+1

= -1 -d * (a2- an *q)/(1 -q) + cnan+1

= -1 -2* (2^n -2) + (2n-1) 2^n

= (2n-1 -2) 2^n -1 -2*(-2)

= (2n-3) 2^n +3

即

2 Tn - Tn = (2n-3) 2^n +3

Tn = (2n-3) 2^n +3