Question

已知等比数列{an}满足a3=12,s3=36.求数列{an}的通项公式;求数列{nan}的前n

Answer 1

a(n) = aq^(n-1),
12 = a(3) = aq^2.

36 = s(3) = a + aq + aq^2 = a + aq + 12,
24 = a(1+q), q不为-1。
a = 24/(1+q).

12 = aq^2 = 24q^2/(1+q),
12(1+q) = 24q^2,
0 = 2q^2 - q - 1 = (2q+1)(q-1),
q = -1/2或q = 1.

q = -1/2时，a=24/(1+q) = 48. a(n) = 48(-1/2)^(n-1).
t(n) = 1*a(1) + 2*a(2) + 3*a(3) + ... + (n-1)a(n-1) + na(n)
= 48 + 2*48(-1/2) + 3*48*(-1/2)^2 + ... + (n-1)48(-1/2)^(n-2) + n48(-1/2)^(n-1)
(-1/2)t(n) = 48(-1/2) + 2*48*(-1/2)^2 + ... + (n-1)48(-1/2)^(n-1) + n48(-1/2)^n

(3/2)t(n) = t(n) - (-1/2)t(n) = 48 + 48(-1/2) + 48(-1/2)^2 + ... + 48(-1/2)^(n-1) - n*48(-1/2)^n.
= 48[1 + (-1/2) + (-1/2)^2 + ... + (-1/2)^(n-1)] - 48n(-1/2)^n
= 48[1 - (-1/2)^n]/(1+1/2) - 48n(-1/2)^n
= 48(2/3)[1- (-1/2)^n] - 48n(-1/2)^n
= 32 - (32 + 48n)(-1/2)^n,

t(n) = 64/3 - (64/3 + 32n)(-1/2)^n

q = 1时，a = 24/(1+q) = 12. a(n) = 12.
t(n) = 1*a(1) + 2*a(2) + 3*a(3) + ... + (n-1)a(n-1) + na(n) = 12[1 + 2 + 3 + ... + n] = 12n(n+1)/2
= 6n(n+1)

Answer 2

因为等比数列a3=a1q^2=12
所以a1＝12/q^2
而s3＝a1（1-q^3）/（1-q）＝36
所以12/q^2*（1-q^3）/（1-q）＝12*(1+q+q^2)/q^2=36
所以2q^2-q-1＝0
所以q＝-1/2或q＝1（舍弃，等比数列公比不等于1）
所以a1＝48
所以an＝48*（-1/2）^(n-1)

所以sn＝48*（1-（-1/2）^n）/（1+1/2）＝2/3*（1-（-1/2）^n）