已知数列{a n }的前n项和为S n 且满足S n = 1 2 n 2 + 1 2 n.数列{b n }满足b 1 =1 2b n -b n-1 =0(n≥

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导读 【#已知数列{a n }的前n项和为S n 且满足S n = 1 2 n 2 + 1 2 n.数列{b n }满足b 1 =1 2b n -b n-1 =0(n≥#】...
【#已知数列{a n }的前n项和为S n 且满足S n = 1 2 n 2 + 1 2 n.数列{b n }满足b 1 =1 2b n -b n-1 =0(n≥#】

(Ⅰ)当n>1时,an=Sn-Sn-1=n,当n=1时,求得a1=s1=1.所以an=n.因为bnbn−1=12且b1=1,所以bn=(12)n−1.…(6分)(Ⅱ)由(Ⅰ),知cn=n•(12)n−1.所以Tn=1•(12)0+2•(12)1+…+n•(12)n−1,12Tn=1•(12)1+2...

【#已知数列{a n }的前n项和为S n 且满足S n = 1 2 n 2 + 1 2 n.数列{b n }满足b 1 =1 2b n -b n-1 =0(n≥#】到此分享完毕,希望对大家有所帮助。

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