已知数列An,A1=1,An=kA(n-1)+k-2,若k=3,令bn=An+1/2,求数列bn的前n项和Sn 谢谢o(∩_∩)o 哈
人气:443 ℃ 时间:2020-06-20 20:12:50
解答
这题简单
k=3,则An=3A(n-1)+3-2=3A(n-1)+1,
将此式子变形为
An+1/2=3A(n-1)+3/2
即
An+1/2=3(A(n-1)+1/2)
即
(An+1/2)/ (A(n-1)+1/2)=3
所以{An+1/2}是以A1+1/2=3/2为首项,3为公比的等比数列,
bn=An+1/2,所以{bn}是以b1=3/2为首项,3为公比的等比数列.
所以Sn=b1(1-q^n)/(1-q)
=(3/2)(1-(3^n))/(1-3)
=(3/4)(3^n-1)
=3^(n+1)/4-3/4
即数列bn的前n项和Sn=3^(n+1)/4-3/4.
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