设数列｛an}满足：a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)

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设数列｛an}满足：a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)
(1)令bn=an+1-an(n=1,2,3,...),求数列｛bn}的通项公式；
（2）求数列{an}的前n项和Sn．
谢拉．．．急ING~

由an+2=5/3an+1-2/3an
可得an+2-an+1=2/3an+1-2/3an
bn+1=2bn/3
bn+1/bn=2/3
｛bn}是公比为2/3的等比数列
b1=a2-a1=2/3
bn=(2/3)^n
设Sn为｛bn}前n项和
Sn=2[1-(2/3)^n]=a2-a1+a3-a2+a4-a3……+an+1-an
=an+1-a1=an+1-1
an+1=2[1-(2/3)^n]+1
an=2[1-(2/3)^(n-1)]+1
an=3-2(2/3)^(n-1)
数列{an}的前n项和Sn
Sn=3n-2[(2/3)^0+(2/3)^1+……+(2/3)^(n-1)]
Sn=3n-6+4(2/3)^(n-2)
计算结果也许有小问题,思路绝对正确

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