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            • 1. 为治疗某种疾病,研制了甲、乙两种新药,希望知道哪种新药更有效,为此进行动物试验.试验方案如下:每一轮选取两只白鼠对药效进行对比试验.对于两只白鼠,随机选一只施以甲药,另一只施以乙药.一轮的治疗结果得出后,再安排下一轮试验.当其中一种药治愈的白鼠比另一种药治愈的白鼠多4只时,就停止试验,并认为治愈只数多的药更有效.为了方便描述问题,约定:对于每轮试验,若施以甲药的白鼠治愈且施以乙药的白鼠未治愈则甲药得1分,乙药得-1分;若施以乙药的白鼠治愈且施以甲药的白鼠未治愈则乙药得1分,甲药得-1分;若都治愈或都未治愈则两种药均得0分.甲、乙两种药的治愈率分别记为α和β,一轮试验中甲药的得分记为X.
              (1)求X的分布列;
              (2)若甲药、乙药在试验开始时都赋予4分,pi(i=0,1,…,8)表示“甲药的累计得分为i时,最终认为甲药比乙药更有效”的概率,则p0=0,p8=1,pi=api-1+bpi+cpi+1(i=1,2,…,7),其中a=P(X=-1),b=P(X=0),c=P(X=1).假设α=0.5,β=0.8.
              (i)证明:{pi+1-pi}(i=0,1,2,…,7)为等比数列;
              (ii)求p4,并根据p4的值解释这种试验方案的合理性.
            • 2.
              已知等比数列\(\{a_{n}\}\)的公比\(q > 1\),且\(a_{3}+a_{4}+a_{5}=28\),\(a_{4}+2\)是\(a_{3}\),\(a_{5}\)的等差中项\(.\)数列\(\{b_{n}\}\)满足\(b_{1}=1\),数列\(\{(b_{n+1}-b_{n})a_{n}\}\)的前\(n\)项和为\(2n^{2}+n\).
              \((\)Ⅰ\()\)求\(q\)的值;
              \((\)Ⅱ\()\)求数列\(\{b_{n}\}\)的通项公式.
            • 3.
              设\(\{a_{n}\}\)是等比数列,公比大于\(0\),其前\(n\)项和为\(S_{n}(n∈N*)\),\(\{b_{n}\}\)是等差数列\(.\)已知\(a_{1}=1\),\(a_{3}=a_{2}+2\),\(a_{4}=b_{3}+b_{5}\),\(a_{5}=b_{4}+2b_{6}\).
              \((\)Ⅰ\()\)求\(\{a_{n}\}\)和\(\{b_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)设数列\(\{S_{n}\}\)的前\(n\)项和为\(T_{n}(n∈N*)\),
              \((i)\)求\(T_{n}\);
              \((ii)\)证明\( \sum\limits_{k=1}^{n} \dfrac {(T_{k}+b_{k+2})b_{k}}{(k+1)(k+2)}= \dfrac {2^{n+2}}{n+2}-2(n∈N*)\).
            • 4.
              给定无穷数列\(\{a_{n}\}\),若无穷数列\(\{b_{n}\}\)满足:对任意\(n∈N^{*}\),都有\(|b_{n}-a_{n}|\leqslant 1\),则称\(\{b_{n}\}\)与\(\{a_{n}\}\)“接近”.
              \((1)\)设\(\{a_{n}\}\)是首项为\(1\),公比为\( \dfrac {1}{2}\)的等比数列,\(b_{n}=a_{n+1}+1\),\(n∈N^{*}\),判断数列\(\{b_{n}\}\)是否与\(\{a_{n}\}\)接近,并说明理由;
              \((2)\)设数列\(\{a_{n}\}\)的前四项为:\(a_{1}=1\),\(a_{2}=2\),\(a_{3}=4\),\(a_{4}=8\),\(\{b_{n}\}\)是一个与\(\{a_{n}\}\)接近的数列,记集合\(M=\{x|x=b_{i},i=1,2,3,4\}\),求\(M\)中元素的个数\(m\);
              \((3)\)已知\(\{a_{n}\}\)是公差为\(d\)的等差数列,若存在数列\(\{b_{n}\}\)满足:\(\{b_{n}\}\)与\(\{a_{n}\}\)接近,且在\(b_{2}-b_{1}\),\(b_{3}-b_{2}\),\(…\),\(b_{201}-b_{200}\)中至少有\(100\)个为正数,求\(d\)的取值范围.
            • 5.
              设\(\{a_{n}\}\)是等差数列,其前\(n\)项和为\(S_{n}(n∈N*)\);\(\{b_{n}\}\)是等比数列,公比大于\(0\),其前\(n\)项和为\(T_{n}(n∈N*).\)已知\(b_{1}=1\),\(b_{3}=b_{2}+2\),\(b_{4}=a_{3}+a_{5}\),\(b_{5}=a_{4}+2a_{6}\).
              \((\)Ⅰ\()\)求\(S_{n}\)和\(T_{n}\);
              \((\)Ⅱ\()\)若\(S_{n}+(T_{1}+T_{2}+……+T_{n})=a_{n}+4b_{n}\),求正整数\(n\)的值.
            • 6.

              \(S\)\({\,\!}_{n}\)为等差数列\(\{a_{n}\}\)的前\(n\)项和,且\({{a}_{1}}\)\(=1\) ,\({{S}_{7}}\)\(=28\)  记\(b_{n}=[\lg a_{n}]\),其中\([x]\)表示不超过\(x\)的最大整数,如\([0.9] = 0\),\([\lg 99]=1\)。

              \((I)\)求\({{b}_{1}}\),\({{b}_{11}}\),\({{b}_{101}}\);

              \((II)\)求数列\(\{b_{n}\}\)的前\(1 000\)项和.

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