共50条信息
已知数列\(\left\{ {{a}_{n}} \right\}\)满足:\({{a}_{1}}=3\),\({{a}_{n+1}}=\dfrac{1}{1-{{a}_{n}}}\),则\({{a}_{2020}}=(\) \()\)
已知数列\(\{{{a}_{n}}\}\)的前\(n\)项和为\({{S}_{n}}\),且\({{a}_{1}}=2\),\({{S}_{n}}-{{S}_{n-1}}=\dfrac{n}{2(n-1)}{{a}_{n-1}}(n\geqslant 2)\).
\((1)\)证明:数列\(\{\dfrac{{{a}_{n}}}{n}\}\)是等比数列,并求通项公式\({{a}_{n}}\) ;
\((2)\)求\({{S}_{n}}\).
已知数列\(\{ a_{n}\}\)的首项为\(a_{1}{=}1\),\(a_{2}{=}3\),且满足对任意的\(n{∈}N^{{⋅}}\),都有\(a_{n{+}1}{-}a_{n}{\leqslant }2^{n}\),\(a_{n{+}2}{-}a_{n}{\geqslant }3{×}2^{n}\)成立,则\(a_{2015}{=}\) ______ .
等差数列\(\{a_{n}\}\)中,\(2a_{1}+3a_{2}=11\),\(2a_{3}+a_{6}-4\),其前\(n\)项和为\(S_{n}\).
\((1)\)求数列\(\{a_{n}\}\)的通项公式;
\((2)\)设数列\(\{b_{n}\}\)满足\({{b}_{n}}=\dfrac{1}{{{S}_{n+1}}-1}\),其前\(n\)项和为\(T_{n}\),求证:\({{T}_{n}} < \dfrac{3}{4}(n∈N^{*})\)
已知\(f(n)=1^{2}+2^{2}+3^{2}+…+(2n)^{2}\),则\(f(k+1)\)与\(f(k)\)的关系是\((\) \()\)
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