求解通项公式

学术版

PrincessQi @ 2020-10-05 09:51:25

b_i=(1-x_2)b_{i-1}+1-(1-x_1)^{i-2}

的通项公式。


by PrincessQi @ 2020-10-05 09:52:57

其中b_2=0


by _biscuitbc @ 2020-10-05 09:54:53

x1,x2是常数么?


by PrincessQi @ 2020-10-05 09:55:40

@Illusory_ 是


by 鏡音リン @ 2020-10-05 10:03:43

解不出来,wtcl


by _biscuitbc @ 2020-10-05 10:11:46

感觉我数学白学了..


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