设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an.求an

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设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an.求an

设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an.求an
设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an.求an

设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an.求an
你写的分数分不清是几,但做题思路是这样,令n=1,带入an+2=5/3an+1-2/3an,得到a3,再令n=2,得到a4,发现规律了没,总结规律,然后加以验证即可

an=3(1-(2/3)的n次方)

由 a(n+2) = 5/3a(n+1) - 2/3an可得,
a(n+2) - a(n+1) = 2/3[a(n+1)-an]
[a(n+2)-a(n+1)] / [a(n+1)-an] = 2/3
设 [a(n+1)-an] = bn
则 b(n+1)/bn = 2/3
b1 = a2 - a1 = 2/3
所以,bn = 2/3 ...

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由 a(n+2) = 5/3a(n+1) - 2/3an可得,
a(n+2) - a(n+1) = 2/3[a(n+1)-an]
[a(n+2)-a(n+1)] / [a(n+1)-an] = 2/3
设 [a(n+1)-an] = bn
则 b(n+1)/bn = 2/3
b1 = a2 - a1 = 2/3
所以,bn = 2/3 *(2/3)^(n-1) = (2/3)^n
所以,a(n+1)-an = (2/3)^n
那么,an-a(n-1) = (2/3)^(n-1)
a(n-1)-a(n-2) = (2/3)^(n-2)
: : :
: : :
a2 - a1 = 2/3
将以上n项相加,得,
a(n+1) - a1 = 2/3*[1-(2/3)^n]/(1-2/3)
a(n+1) - 1 = 2*[1-(2/3)^n]
a(n+1) = 3 - 2*(2/3)^n
即,an = 3 - 2*(2/3)^(n-1)
那所谓的满意回答 太扯了吧
求数列是有方法的 这个数列不是规律数列 所以不能一一求a3 ,a4

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