用反证法证明数学命题时，首先应该做出与命题结论相反的假设，否定“自然数\(a\ ,\ b\ ,\ c\)中恰有一个偶数”时正确的反设为                          \((\)   \()\)

已知点\({{P}_{n}}({{a}_{n}},{{b}_{n}})(n\in {{N}^{+}})\)在直线\(l:y=3x+1\)上，\({{P}_{1}}\)是直线\(l\)与\(y\)轴的交点，数列\(\left\{ {{a}_{n}} \right\}\)是公差为\(1\)的等差数列．

\((\)Ⅰ\()\)求数列\(\left\{ {{a}_{n}} \right\}\)，\(\left\{ {{b}_{n}} \right\}\)的通项公式；

\((\)Ⅱ\()\)求证：\(\dfrac{1}{{{\left| {{P}_{1}}{{P}_{2}} \right|}^{2}}}+\dfrac{1}{{{\left| {{P}_{1}}{{P}_{3}} \right|}^{2}}}+......+\dfrac{1}{{{\left| {{P}_{1}}{{P}_{n+1}} \right|}^{2}}} < \dfrac{1}{5}\)

如果无穷数列\(\{{{a}_{n}}\}\)满足下列条件：\(①\dfrac{{{a}_{n}}+{{a}_{n+2}}}{2}\leqslant {{a}_{n+1}}\)；\(②\)存在实数\(M\)，使得\({{a}_{n}}\leqslant M\)，其中\(n\in {{N}^{*}}\)，那么我们称数列\(\{{{a}_{n}}\}\)为\(\Omega \)数列．

\((1)\)设数列\(\{{{b}_{n}}\}\)的通项为\({{b}_{n}}=5n-{{2}^{n}}\)，且是\(\Omega \)数列，求\(M\)的取值范围；

\((2)\)设\(\{{{c}_{n}}\}\)是各项为正数的等比数列，\({{S}_{n}}\)是其前\(n\)项和，\({{c}_{3}}=\dfrac{1}{4},{{S}_{3}}=\dfrac{7}{4}\)，证明：数列\(\{{{S}_{n}}\}\)是\(\Omega \)数列；

\((3)\)设数列\(\{{{d}_{n}}\}\)是各项均为正整数的\(\Omega \)数列，求证：\({{d}_{n}}\leqslant {{d}_{n+1}}\)．

\((1)\)求证：\(1+ \dfrac{1}{{2}^{2}}+ \dfrac{1}{{3}^{2}}+⋯+ \dfrac{1}{{n}^{2}} < 2- \dfrac{1}{n} (n∈N^{*},n\geqslant 2)\)

\((2)\)设\(a\)，\(b\)，\(c∈R\)，证明：\(a^{2}+b^{2}+c^{2}\geqslant ab+ac+bc\)．

如果非零实数\(a,b,c \) 两两不相等，且\(2b=a+c .\)证明：\( \dfrac{2}{b}\neq \dfrac{1}{a}+ \dfrac{1}{c} \)．

正数数列\(\left\{{a}_{n}\right\} \)、\(\left\{{b}_{n}\right\} \)满足：\({a}_{1}\geqslant {b}_{1} \)，且对一切\(k\geqslant 2,k∈{N}^{*} \)，\({a}_{k} \)是\({a}_{k-1} \)与\({b}_{k-1} \)的等差中项，\({b}_{k} \)是\({a}_{k-1} \)与\({b}_{k-1} \)的等比中项．

\((1)\)若\({a}_{2}=2,{b}_{2}=1 \)，求\({a}_{1},{b}_{1} \)的值；

\((2)\)求证：\(\left\{{a}_{n}\right\} \)是等差数列的充要条件是\(\left\{{a}_{n}\right\} \)为常数数列；

\((3)\)记\({c}_{n}=\left|{a}_{n}-{b}_{n}\right| \)，当\(n\geqslant 2\left(n∈{N}^{*}\right) \)时，指出\({c}_{2}+⋯+{c}_{n} \)与\({c}_{1} \)的大小关系并说明理由．

设各项均为正数的数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且\(S_{n}\)满足\(S_{n}^{2}-\left( {{n}^{2}}+n-3 \right){{S}_{n}}-3\left( {{n}^{2}}+n \right)=0\) ，\(n∈N*\)．

\((1)\)求\(a_{1}\)的值；

\((2)\)求数列\(\{a_{n}\}\)的通项公式；

\((3)\)证明：对一切正整数\(n\)，有\(\dfrac{1}{{{a}_{1}}\left( {{a}_{1}}+1 \right)}+\dfrac{1}{{{a}_{2}}\left( {{a}_{2}}+1 \right)}+\cdots +\dfrac{1}{{{a}_{n}}\left( {{a}_{n}}+1 \right)} < \dfrac{1}{3}\) ．