数列1，\(1 + 2\)，\(1 + 2 + 4\)，…，\(1 + 2 + 4 + \cdots + {2^n}\)，…的前\(n\)项和为（       ）．

设 \({a_n} = 1 + 2 + 4 + \cdots + {2^{n - 1}} = \frac{{1 \cdot \left( {1 - {2^n}} \right)}}{{1 - 2}} = {2^n} - 1\)，

所以数列 \(\left\{ {{a_n}} \right\}\) 的前 n 项和 \({T_n} = {a_1} + {a_2} + \cdot \cdot \cdot + {a_n} = 2 - 1 + {2^2} - 1 + \cdot \cdot \cdot + {2^n} - 1\)

\( = \left( {2 + {2^2} + \cdot \cdot \cdot + {2^n}} \right) - n = \frac{{2\left( {1 - {2^n}} \right)}}{{1 - 2}} - n = {2^{n + 1}} - 2 - n\).

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