已知{an)为等比数列,其前n项和为Sn,且Sn=2^n+a

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 +3bn=(2n-1)an=(2n-1)2^(n-1) Tn=1.2^0+3.2^1+5.2^2+...+2(n-1)(2n-1)2Tn=1.2^1+3.2^2+5.2^3+...+(2n-3)(2^n-1)+(2n-1)2^nTn-2Tn=1+2.2^1+2.2^2+...+(2.2^n-1)-(2n-1)2^n1-(2n-1)2^n+2(2^1+2^2+...+2^n-1）=（3-2n)(2^n)-3我这样做为什么答案不对啊？Tn - 2Tn = -Tn; 结果取相反数好了; -[ (3-2n)(2^n)-3] = -(3-2n)(2^n)+3 =(2n-3)(2^n)+3 易于验证 T1 = b1 = c1 a1 = 1*1 T1 = (2*1-3)*2^1 +3 = -1*2 +3 = 1结果是对的