数列 \({a_1},{a_2} - {a_1},{a_3} - {a_2}, \ldots ,{a_n} - {a_n}_{ - 1}\) ，…是首项为\(1\)，公比为\(2\)的等比数列，那么\({a_n} = \) ___.

\({a_n} - {a_n}_{ - 1} = {a_1}{q^n}^{ - 1} = {2^n}^{ - 1}\) ，

即 \(\left\{ {\begin{array}{*{20}{l}}
{{a_2} - {a_1} = 2,} \\\
{{a_3} - {a_2} = {2^2},} \\\
{...} \\\
{{a_n} - {a_{n - 1}} = {2^{n - 1}}.}
\end{array}} \right.\) 

各式相加得 \({a_n} - {a_1} = 2 + {2^2} + \ldots + {2^n}^{ - 1} = {2^n} - 2\) ，

故 \({a_n} = {a_1} + {2^n} - 2 = {2^n} - 1\) .

故答案为： \({2^n} - 1\)

进入考试题库