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            • 1.
              数列\(\{a_{n}\}\)满足\(a_{n}=4a_{n-1}+3(n\geqslant 2\)且\(n∈N*)\),\(a_{1}=1\),则此数列的第\(3\)项是\((\)  \()\)
              A.\(15\)
              B.\(255\)
              C.\(20\)
              D.\(31\)
            • 2.
              已知数列\(\{a_{n}\}\)的通项\(a_{n}=2^{n}- \dfrac {1}{2}(n+3)\),若数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),则\(S_{8}=\) ______ \(.(\)用数字作答\()\)
            • 3.
              设数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),若\(S_{n}=n^{2}\),则\(a_{8}=\) ______ .
            • 4.
              已知\(S_{n}\)为数列\(\{a_{n}\}\)的前\(n\)项和,且\(\log _{2}(S_{n}+1)=n+1\),则数列\(\{a_{n}\}\)的通项公式为\((\)  \()\)
              A.\(a_{n}=2^{n}\)
              B.\(a_{n}= \begin{cases} \overset{3\;n=1}{2^{n}n\geqslant 2}\end{cases}\)
              C.\(a_{n}=2^{n-1}\)
              D.\(a_{n}=2^{n+1}\)
            • 5.
              已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}= \dfrac {n^{2}+n}{2}\),\(n∈N^{*}\).
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)设\(b_{n}=2a_{n}+(-1)^{n}a_{n}\),求数列\(\{b_{n}\}\)的前\(2n\)项和.
            • 6.

              已知数列\(\{ a_{n}\}\)的前\(n\)项和\(S_{n}{=}n^{2}\),则\(a_{3}{-}a_{2}\)的值为\((\)  \()\)

              A.\({-}2\)
              B.\(2\)
              C.\({-}3\)
              D.\(3\)
            • 7.

              在数列\(\{a_{n}\}\)中,\(a_{1}=1\),\({{a}_{n}}=1+\dfrac{{{\left( -1 \right)}^{n}}}{{{a}_{n-1}}}(n\geqslant 2)\),则\(a_{5}\)等于

              A.\(\dfrac{3}{2}\)
              B.\(\dfrac{5}{3}\)
              C.\(\dfrac{8}{5}\)
              D.\(\dfrac{2}{3}\)
            • 8.

              已知数列\(\{a_{n}\}\)满足\(S_{n}+a_{n}=2n+1\),\((n∈N*)\)

              \((\)Ⅰ\()\)求出\(a_{1}\),\(a_{2}\),\(a_{3}\)的值;

              \((\)Ⅱ\()\)由\((\)Ⅰ\()\)猜测\(a_{n}\)的表达式,并用数学归纳法证明所得结论.

            • 9. 已知数列\(\left\{ {{a}_{n}} \right\}\)中,\({{a}_{1}}=1\),\({{a}_{n+1}}=\dfrac{2{{a}_{n}}}{2+{{a}_{n}}}(n\in {{N}^{*}})\),则可归纳猜想\(\left\{ {{a}_{n}} \right\}\)的通项公式为(    )
              A.\({{a}_{n}}=\dfrac{2}{n}\)
              B.\({{a}_{n}}=\dfrac{2}{n+1}\)
              C.\({{a}_{n}}=\dfrac{1}{n}\)
              D.\({{a}_{n}}=\dfrac{1}{n+1}\)
            • 10.

              已知数列\(\left\{{a}_{n}\right\} \)的前\(n \)项和为\({S}_{n} \),\({a}_{1}=2 \),\({a}_{n+1}-{S}_{n}=2\left(n∈{N}^{*}\right) \),则\({a}_{n}= \)       

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