已知数列{an}是等差数列 a4=7 S3=9 数列前n项和为Sn 且Sn=2bn-2 求{an}{bn}的通项公式
问题描述:
已知数列{an}是等差数列 a4=7 S3=9 数列前n项和为Sn 且Sn=2bn-2 求{an}{bn}的通项公式
求 数列{an bn}的前n项Tn
答
设{an}等差为x,则
a1 + 3x = 7
3a1 + 3x = 9
则:
a1 = 1
x = 2
则:
an = 1 + 2(n-1) = 2n -1
则
sn = (a1 + an) *n /2 = n^2
则:
2bn-2 =sn = n^2
则
bn = (n^2 + 2)/2