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已知公差为正数的等差数列
求数列
参考答案:解:设公差为 \(d > 0\),公比为 \(q\),由题,因为 \(\left\{ {\begin{array}{*{20}{l}}
{{a_2} = {b_2}} \\
{{a_3} + 1 = {b_3}}
\end{array}} \right.\),则 \(\left\{ {\begin{array}{*{20}{l}}
{1 + d = 1 \cdot q} \\
{1 + 2d + 1 = 1 \cdot {q^2}}
\end{array}} \right.\),解得 \(\left\{ {\begin{array}{*{20}{l}}
{d = 1} \\
{q = 2}
\end{array}} \right.\),所以 \({a_n} = 1 + \left( {n - 1} \right) \times 1 = n\),\({b_n} = 1 \times {2^{n - 1}} = {2^{n - 1}}\).