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

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

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

已知数列{an}满足：a1=1,a(n+1)=2an+1(n∈N*),用数学归纳法证明：an=2^n-1RT
证明：
n=1时,a1=1
假设n=k(k为自然数,且k>1)时,
ak=2^k-1
则当n=k+1时,
a(k+1)=2ak+1=2*(2^k-1)+1=2^(k+1)-1
综上,得an=2^n-1

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