设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an,(n属于N※),求数列(bn)的通项公式.(2)求数列{nan}的前n项和Sn

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设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an,(n属于N※),求数列(bn)的通项公式.(2)求数列{nan}的前n项和Sn

设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an,(n属于N※),求数列(bn)的通项公式.(2)求数列{nan}的前n项和Sn
设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).
(1)令bn=an+1-an,(n属于N※),求数列(bn)的通项公式.(2)求数列{nan}的前n项和Sn

设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an,(n属于N※),求数列(bn)的通项公式.(2)求数列{nan}的前n项和Sn
(I)因bn=an+1-an,所以bn+1=an+2-an+1=5/3an+1-2/3an-an+1=2/3(an+1-an)=2/3bn
故{bn}是公比为2/3的等比数列,且b1=a2-a1=2/3
\x09(II)由 bn=an+1-an
得an+1-a1=(an+1-an)+(an-an-1)+……+(a2-a1)
=(2/3)n次方+(2/3)n-1次方+……+(2/3)
设Tn={n2n-1次方/3n-1次方}
a1=1,得an=3-2n次方/3n-1次方
Tn-1/3Tn=9-(3+n)2n次方/3n-1次方
Sn=a1+2a2+……+nan=3/2n(n+1)+(3+n)2n次方+1/3n-1次方 -18

可以看出B1=A2-A1=2/3
Bn+1=An+2-An+1=5/3An+1 - 2/3An
可以得An=Bn+1-5/3Bn
代入Bn=Bn+2 - 5/3Bn+1 - Bn+1 + 5/3Bn = Bn+2 - 8/3Bn+1 + Bn
话说此题打字也很烦~~
的确悬赏分有点少

这种题悬赏分也太少啦

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