题目内容
设an=1+q+q2+…+qn-1(n∈N,q≠±1),An=C n1a1+C n2a2+…+Cnnan,求An(用n和q表示).
An= [2n-(1-q)n]
解析解:因为an=,
所以An= [C n1 (1-q)+C n2 (1-q2)+…+Cnn (1-qn)]
= [C n1+C n2+…+Cnn-(Cn1q+Cn2q2+…+Cnnqn)]
= [(2n-1)-(1+q)n+1]
= [2n-(1-q)n].
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