已知数列\(\{{{a}_{n}}\}\)
满足\({{a}_{1}}=\sqrt{2}\)
,\(a_{n}^{2}-a_{n-1}^{2}=2n\left(n\geqslant 2\right) \)
,且\({{a}_{n}} > 0\)
. \((1)\)求\(\{{{a}_{n}}\}\)
的通项; \((2)\)设\(\{{{a}_{n}}\}\)的前\(n\)项和为\({{S}_{n}}\),用数学归纳法证明:\({{S}_{n}} < \dfrac{1}{2}{{(n+1)}^{2}}\).