已知数列an满足a1=7/8,且an+1=1/2an+1/3,n是正整数,求an通项公式
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已知数列an满足a1=7/8,且an+1=1/2an+1/3,n是正整数,求an通项公式
已知数列an满足a1=7/8,且an+1=1/2an+1/3,n是正整数,求an通项公式
已知数列an满足a1=7/8,且an+1=1/2an+1/3,n是正整数,求an通项公式
a(n+1)=(1/2)an +1/3
a(n+1)-2/3=(1/2)an -1/3=(1/2)(an -2/3)
[a(n+1) -2/3]/(an -2/3)=1/2,为定值.
a1 -2/3=7/8 -2/3=5/24,数列{an -2/3}是以5/24为首项,1/2为公比的等比数列.
an -2/3=(5/24)×(1/2)^(n-1)=(5/3)×(1/2)^(n+2)
an=[5/2^(n+2) +2]/3
n=1时,a1=(5/2^3 +2)/3=(5/8 +2)/3=7/8,同样满足通项公式
数列{an}的通项公式为an=[5/ 2^(n+2) +2]/3
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