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用反证法证明“三角形中最多只有一个内角为钝角”,下列假设中正确的是( )
证明:对于\(n∈N^{*}\),不等式\(|\sin nθ|\leqslant n|\sin θ|\)恒成立.
已知数列\(\{a_{n}\}\)满足\(a_{1}=1\),\(a_{n+1}=3a_{n}+1\).
\((1)\)证明\(\left\{ {{a}_{n}}+\dfrac{1}{2} \right\}\)是等比数列,并求\(\{a_{n}\}\)的通项公式;
\((2)\)证明:\(\dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+\cdots +\dfrac{1}{{{a}_{n}}} < \dfrac{3}{2}\).
求证:\( \dfrac{1}{1^{2}}+ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+…+ \dfrac{1}{n^{2}} < 2\).
设\(n\)是正整数,求证:\(\dfrac{1}{2} \leqslant \dfrac{1}{n+1} +\dfrac{1}{n+2} +…+\dfrac{1}{2n} < 1\).
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