数列 $\left\{ {{a_n}} \right\}$ 满足 $a_1=1$，前 $n$ 项和为 ${S_n}$，${S_{n + 1}} = 4{a_n} + 2$，则 ${a_{2013}}=$ \((\qquad)\)
【难度】
【出处】
2013年北京大学等三校联考自主招生保送生测试
【标注】
【答案】
A
【解析】
由 ${a_1} + {a_2} = 4{a_1} + 2$，得 ${a_2} = 5$．
${S_{n + 1}} = 4{a_n} + 2$，于是$${a_{n + 2}} = 4\left( {{a_{n + 1}} - {a_n}} \right),$$即$${a_{n + 2}} - 2{a_{n + 1}} = 2\left( {{a_{n + 1}} - 2{a_n}} \right).$$所以$${a_{n + 1}} - 2{a_n} = 3 \cdot {2^{n - 1}},$$从而$$\dfrac{{{a_{n + 1}}}}{{{2^{n + 1}}}} = \dfrac{{{a_n}}}{{{2^n}}} + \dfrac{3}{4},$$因此$$\dfrac{{{a_n}}}{{{2^n}}} = \dfrac{3}{4}\left( {n - 1} \right) + \dfrac{1}{2},$$解得 ${a_n} = \left( {3n - 1} \right) \cdot {2^{n - 2}}$，于是 ${a_{2013}} = 3019 \cdot {2^{2012}}$．
${S_{n + 1}} = 4{a_n} + 2$，于是$${a_{n + 2}} = 4\left( {{a_{n + 1}} - {a_n}} \right),$$即$${a_{n + 2}} - 2{a_{n + 1}} = 2\left( {{a_{n + 1}} - 2{a_n}} \right).$$所以$${a_{n + 1}} - 2{a_n} = 3 \cdot {2^{n - 1}},$$从而$$\dfrac{{{a_{n + 1}}}}{{{2^{n + 1}}}} = \dfrac{{{a_n}}}{{{2^n}}} + \dfrac{3}{4},$$因此$$\dfrac{{{a_n}}}{{{2^n}}} = \dfrac{3}{4}\left( {n - 1} \right) + \dfrac{1}{2},$$解得 ${a_n} = \left( {3n - 1} \right) \cdot {2^{n - 2}}$，于是 ${a_{2013}} = 3019 \cdot {2^{2012}}$．
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