已知 $\left\{ {a_n}\right\} $ 是各项均为正数的等比数列，且 ${a_1} + {a_2} = 2\left(\dfrac{1}{a_1} + \dfrac{1}{a_2}\right)$，${a_3} + {a_4} + {a_5} = 64\left(\dfrac{1}{a_3} + \dfrac{1}{a_4} + \dfrac{1}{a_5}\right)$．
【难度】
【出处】
2010年高考大纲全国II卷（文）
【标注】
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求 $\left\{ {a_n}\right\} $ 的通项公式；标注答案解析设公比为 $ q $，则 ${a_n} = {a_1}{q^{n - 1}}$．由已知有\[{\begin{cases}
{a_1} + {a_1}q = 2\left( {\dfrac{1}{a_1} + \dfrac{1}{{{a_1}q}}} \right), \\
{a_1}{q^2} + {a_1}{q^3} + {a_1}{q^4} = 64\left( {\dfrac{1}{{{a_1}{q^2}}} + \dfrac{1}{{{a_1}{q^3}}} + \dfrac{1}{{{a_1}{q^4}}}} \right). \\
\end{cases}}\]化简得\[{\begin{cases}{a_1}^2q = 2, \\
{a_1}^2{q^6} = 64. \\
\end{cases}}\]又 $a_1>0 $，故 $q=2 $，$ a_1=1$．所以 $a_n=2^{n-1} $． -
设 ${b_n} = {\left({a_n} + \dfrac{1}{a_n}\right)^2}$，求数列 $\left\{ {b_n}\right\} $ 的前 $n$ 项和 ${T_n}$．标注答案解析由（1）知\[ b_n=\left(a_n+\dfrac 1 {a_n} \right)^2 = a_n^2 +\dfrac 1 {a_n^2} +2 = 4^{n-1} +\dfrac 1 {4^{n-1}} +2 . \]因此\[ \begin{split}T_n & =\left(1+4+\cdots +4^{n-1}\right) + \left(1+\dfrac 1 4 +\cdots +\dfrac 1 {4^{n-1}} \right) +2n \\& = \dfrac {4^n-1} {4-1} +\dfrac {1-\dfrac 1 {4^n} } {1-\dfrac 1 4 } +2n \\&= \dfrac 1 3 \left(4^n - 4^{1-n}\right) +2n+1 .\end{split}\]
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