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            • 1.

              各项都为正数的数列\(\{a_{n}\}\),其前\(n\)项的和为\(S_{n}\),且\(S_{n}=(\sqrt{{S}_{n-1}}+ \sqrt{{a}_{1}} )^{2}(n\geqslant 2)\),若\(b_{n}=\dfrac{{a}_{n+1}}{{a}_{n}}+ \dfrac{{a}_{n}}{{a}_{n+1}} \),且数列\(\{b_{n}\}\)的前\(n\)项的和为\(T_{n}\),则\(T_{n}=\)_____________.

            • 2.

              \(《\)九章算术\(》\)中的“竹九节”问题:现有一根\(9\)节的竹子,自上而下各节的容积成等差数列,上面\(4\)节的容积共\(3L\),下面\(3\)节的容积共\(4L\),则第\(5\)节的容积为              L.

            • 3. 在等差数列{an}中,a1=3,d=2.an=25,则n= ______
            • 4. 等差数列8,5,2,…的前20项和是(  )
              A.410
              B.-410
              C.49
              D.-49
            • 5.
              已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),\(a_{1}=2\),\(S_{n}=n^{2}+n\).
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)设\(\{ \dfrac {1}{S_{n}}\}\)的前\(n\)项和为\(T_{n}\),求证\(T_{n} < 1\).
            • 6.
              已知等比数列\(\{a_{n}\}\)中,\(a_{1}=3\),\(a_{4}=24\),
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)设等差数列\(\{b_{n}\}\)中,\(b_{2}=a_{2}\),\(b_{9}=a_{5}\),求数列\(\{b_{n}\}\)的前\(n\)项和\(S_{n}\).
            • 7.
              已知\(\{a_{n}\}\)是递增的等差数列\(a_{3}= \dfrac {5}{2}\),且\(a_{2}a_{4}=6\).
              \((1)\)求\(\{a_{n}\}\)的首项\(a_{1}\)和公差\(d\);
              \((2)\)求\(\{a_{n}\}\)的通项和前\(n\)项和\(S_{n}\).
            • 8.

              已知数列\({a_{n}}\)的通项公式为\({a}_{n}= \dfrac{1}{1+2+3+…+n} \),

              \((1)\)求\({a}_{1},{a}_{2},{a}_{3} \);    

              \((2)\)求这个数列的前\(n\)项和\(;\)

            • 9. 已知数列\(\{ \)\(a_{n}\)\(\}\)中, \(a\)\({\,\!}_{2}=\) \(a\)\(( \)\(a\)为非零常数\()\),其前 \(n\)项和\(S\) \({\,\!}_{n}\)满足:\(S\) \({\,\!}_{n}\)\(= \dfrac{n\left({a}_{n}-{a}_{1}\right)}{2}\left(n∈{N}^{*}\right) \)
              \((1)\)求数列\(\{ \)\(a_{n}\)\(\}\)的通项公式;
              \((2)\)若 \(a\)\(=2\),且\( \dfrac{1}{4} \) \(a_{m}\)\({\,\!}^{2}-S\) \({\,\!}_{n}\)\(=11\),求 \(m\)\(n\)的值;
            • 10.

              已知在数列\(\{a_{n}\}\)中,其前\(n\)项和\(S_{n}\)满足\({{S}_{n}}=2{{a}_{n}}-4\).

              \((\)Ⅰ\()\) 求数列\(\{a_{n}\}\)的通项公式\(a_{n}\);

              \((\)Ⅱ\()\) 设\({{b}_{n}}=\dfrac{{{a}_{n}}}{n(n+1)\cdot {{2}^{n}}}\),求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\).

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