已知数列{an}满足a1=1,a(n+1)=2an+1 用数学归纳法证明an=2^n-1

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$已知数列{an}满足a1=1,a(n+1)=2an+1 用数学归纳法证明an=2^n-1$

已知数列{an}满足a1=1,a(n+1)=2an+1 用数学归纳法证明an=2^n-1
已知数列{an}满足a1=1,a(n+1)=2an+1 用数学归纳法证明an=2^n-1

已知数列{an}满足a1=1,a(n+1)=2an+1 用数学归纳法证明an=2^n-1
1、显然,当n=1时,an=2^n-1成立
2、下面证明当n=k时成立时,n=k+1也成立
ak=2^k-1
所以ak+1=2*ak+1=2^(k+1)-1
故n=k+1时原式也成立
综上所述,an=2^n-1

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