数列{an}的前n项和Sn满足 sn=32an−12a1 ，且a1，a2+6，a3成等差数列．（Ⅰ）求数列{an}的通项公式；（Ⅱ）设bn= an+1snsn+1 ，求数列{bn}的前n项和Tn．

题干

数列{a_n}的前n项和S_n满足 sn=

3

2

an−

1

2

a1 ，且a₁，a₂+6，a₃成等差数列．

（Ⅰ）求数列{a_n}的通项公式；

（Ⅱ）设b_n=

a

n

+

1

s

n

s

n

+

1

，求数列{b_n}的前n项和T_n．

上一题下一题 0.0难度选择题更新时间:2019-09-22 02:48:21

解：（Ⅰ）由 $s_{n} = \frac{3}{2} a_{n} - \frac{1}{2} a_{1} <$

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