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

n项和Tn

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

追答

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

追问

哥哥 你这是错位相减法吗？

追答

是吧、、其实我也记不清叫什么名字、就知道这样做