1-(1-p)^n>p^n (0
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1-(1-p)^n>p^n(01-(1-p)^n>p^n(01-(1-p)^n>p^n(0证明:设f(n)=p^n+(1-p)^n,因为0<p<1,所以0<1-p<1,所以函数f(x)=p^x+(1-
1-(1-p)^n>p^n (0
1-(1-p)^n>p^n (0
1-(1-p)^n>p^n (0
证明:设f(n)=p^n+(1-p)^n,因为0<p<1,所以0<1-p<1,所以函数f(x)=p^x+(1-p)^x在(0,+∞)上是减函数,于是f(n)≤f(1)=p+1-p=1,即p^n+(1-p)^n≤1,即1-(1-p)^n≥p^n.