记x=(1+2)(1+2 2 )(1+2 4 )(1+2 8 )…(1+2 n ),且x+1=2 128 ,则n=______

2025-01-02 10:48:43
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回答1:

(1+2)(1+2 2 )(1+2 4 )(1+2 8 )…(1+2 n ),
=(2-1)(1+2)(1+2 2 )(1+2 4 )(1+2 8 )…(1+2 n ),
=(2 2 -1)(1+2 2 )(1+2 4 )(1+2 8 )…(1+2 n ),
=(2 n -1)(1+2 n ),
=2 2n -1,
∴x+1=2 2n -1+1=2 2n
2n=128,
∴n=64.
故填64.