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数列前n项和为Sn,a1=1,S(n +1)=4(an)+2.设bn=an/(2的n次方),证:(bn)是等差数列.
人气:386 ℃ 时间:2019-08-19 07:57:59
解答
S(n +1)=4(an)+2
Sn =4(an-1)+2
两式相减
(an+1)=4(an)-4(an-1)
(an+1)+4(an-1)=4(an)
(an+1)/2^(n+1)+4(an-1)/2(n+1)=4(an)/2^(n+1)
(bn+1)+(bn-1)=2(bn)
(bn)是首项为1/2的等差数列
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