数列a1=1,an=2an-1+1(n≥2,n属于N^*)的第5项是?
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数列a1=1,an=2an-1+1(n≥2,n属于N^*)的第5项是?数列a1=1,an=2an-1+1(n≥2,n属于N^*)的第5项是?数列a1=1,an=2an-1+1(n≥2,n属于N^*)的
数列a1=1,an=2an-1+1(n≥2,n属于N^*)的第5项是?
数列a1=1,an=2an-1+1(n≥2,n属于N^*)的第5项是?
数列a1=1,an=2an-1+1(n≥2,n属于N^*)的第5项是?
an=2a(n-1)+1
an+1=2a(n-1)+2
an+1=2[a(n-1)+1]
(an+1)/[a(n-1)+1]=2
所以an+1是以2为公比的等比数列
an+1=(a1+1)*q^(n-1)
an+1=(1+1)*2^(n-1)
an+1=2^n
an=2^n-1
a5=2^5-1=32-1=31
an+1=2(an-1+1)
an+1=2^(n-1) * (a1+1)
an=2^n-1
a5=2^5-1=31
a1=1
a2=2a1+1=3
a3=2a2+1=7
a4=2a3+1=15
a5=2a4+1=31
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