高一数列求高手详解

来源：百度知道 编辑：UC知道 时间：2024/05/26 16:30:00
{an}是公差为1的等差数列,{bn}是公比为2的等比数列,Pn,Qn分别是{an},{bn}的前n项和,且a6=b3,P10=Q4+45
(1)求{an}的通项公式
(2)若Pn＞b6,求n的取值范围

an = a1 + (n-1)
a6 = a1 +5
P10 = (a1 + a10)*10/2 = 5(2a1 + 9) = 10a1 + 45

bn = b1 * 2^(n-1)
b3 = b1 * 2^2 = 4b1
Q4 = b1 * (q^4 -1)/(q-1) = 15 b1

a1 + 5 = 4b1
10a1 + 45 = 15b1 + 45

a1 = 3
b1 = 2

(1){an}的通项公式
an = a1 + (n-1)*1 = 3 + (n-1) = n + 2

(2)若Pn＞b6,求n的取值范围
Pn = (a1 + an)*n/2 = (3 + n+2)*n/2 = n(n+5)/2
b6 = b1 * 2^5 = 2 * 2^5 = 64

n(n+5)/2 > 64
n(n+5) > 128

以 n = 9 和 10 试探
9*(9+5) = 126
10*(10+5) = 150

因此 n的取值范围是 n > 9