已知等差数列 $\left\{ {a_n}\right\} $ 满足：${a_3} = 7$，${a_5} + {a_7} = 26$．数列 $\left\{ {a_n}\right\} $ 的前 $n$ 项和为 ${S_n}$．
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求 ${a_n}$ 及 ${S_n}$；标注答案解析设等差数列 $\left\{ {a_n}\right\} $ 的首项为 ${a_1}$，公差为 $d$，由于\[{a_3} = 7,{a_5} + {a_7} = 26,\]所以\[{a_1} + 2d = 7,2{a_1} + 10d = 26,\]解得\[{a_1} = 3,d = 2.\]由于\[{a_n} = {a_1} + \left(n - 1\right)d,{S_n} = \dfrac{{n\left({a_1} + {a_n}\right)}}{2},\]所以\[{a_n} = 2n + 1,{S_n} = n\left(n + 2\right).\]
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令 ${b_n} = \dfrac{1}{a_n^2 - 1}$ $\left(n \in {{\mathbb{N}}^*}\right)$，求数列 $\left\{ {b_n}\right\} $ 的前 $n$ 项和 ${T_n}$．标注答案解析因为 ${a_n} = 2n + 1$，所以\[a_n^2 - 1 = 4n\left(n + 1\right),\]因此\[{b_n} = \dfrac{1}{4n\left(n + 1\right)} = \dfrac{1}{4}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right).\]故\[\begin{split}{T_n} &= {b_1} + {b_2} + \cdots + {b_n}\\& = \dfrac{1}{4}\left(1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \cdots + \dfrac{1}{n} - \dfrac{1}{n + 1}\right)\\& = \dfrac{1}{4}\left(1 - \dfrac{1}{n + 1}\right)\\& = \dfrac{n}{4\left(n + 1\right)},\end{split}\]所以数列 $\left\{ {b_n}\right\} $ 的前 $n$ 项和 ${T_n} = \dfrac{n}{4\left(n + 1\right)}$．
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