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问题 解答题
已知数列an满足a1=1，n≥2时，
an
an-1
=
2-3an
an-1+2
．
（1）求证：数列{
1
an
}为等差数列；
（2）求{
3n
an
}的前n项和．
答案

（1）证明：由已知

an
an-1
=
2-3an
an-1+2
．

整理可得an-1-an=2an-1an（n≥2），

同时除以anan-1可得

1
an
-
1
an-1
=2，

所以{

1
an
}为首项为
1
a1
=1，公差为2的等差数列．

（2）由（1）可知，

1
an
=1+2(n-1)=2n-1，

所以

3n
an
=(2n-1)3n，

Sn=1×3+3×32+…+（2n-1）×3n①

3Sn=1×32+3×33+…+（2n-1）×3n+1②

①-②得-2Sn=3+2×（32+33+…+3n）-（2n-1）×3n+1=（2-2n）•3n+1-6

所以得Sn=（n-1）3n+1+3

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Jasmine said she wouldn't use such h__小题3:__ products(产品) again. She began to look for natural products on the I__小题4:__. She started mixing her own hair oils and shampoos(洗发水).

In l__小题5:__ than a year, Jasmine had found the wonderful combination (混合物). Her hair started to come back. Soon, friends and family were asking about what she used. Jasmine k__小题6:__ she had a hot product.

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单项选择题

评估某宗房地产2005年9月末的价格，选取的可比实例成交价格为3000元/㎡，成交日期为2005年1月末，该类房地产自2005年1月末至2005年9月末的价格每月与上月的变动幅度为1.5%，2.5%，0.5%，-1.5%，-2.5%，-1.0%，1.5%，-1.5%，0%。则该可比实例在2005年9月末的价格为()元/㎡。

A.2938

B.2982

C.3329

D.3379
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