(2014•深圳一模)已知等差数列{an}中,a1+a3+a5=21,a2+a4...

河北自主招生2022-02-18 05:48:14admin2

解答:解:(1)设等差数列{an}的公差为d,则∴a1+a3+a5=21,a2+a4+a6=27,∴3a3=21,3a4=27,∴a3=7,a4=9,∴d=2,∴an=a3+2(n-3)=2n+1,∴a1=3,∴4Sn=3bn-3,①n=1时,4S1=3b1-3,∴b1=-3,n≥2时,4Sn-1=3bn-1-3②,∴①-②整理得bn=-3bn-1,∴数列{bn}是以-3为首项,-3为公比的等比数列,∴bn=(-3)n;(2)cn=4bn+1bn-1=4+5(-3)n-1,n为奇数时,cn=4-53n+1,∵3n+1≥4,(n=1时取等号)∴114≤4-53n+1<4,n为偶数时,cn=4+53n-1,∵3n-1≥8,(n=2时取等号)∴4<4+53n-1≤378,综上,114≤cn≤378,cn≠4,∴cn=4bn+1bn-1的最小值114,最大值是378.

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