已知数列{an}满足3(an+1)+an=4(n>=1),且a1=9,其前n项和为Sn,则满足不等式{Sn-n-6}
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已知数列{an}满足3(an+1)+an=4(n>=1),且a1=9,其前n项和为Sn,则满足不等式{Sn-n-6}
已知数列{an}满足3(an+1)+an=4(n>=1),且a1=9,其前n项和为Sn,则满足不等式{Sn-n-6}
已知数列{an}满足3(an+1)+an=4(n>=1),且a1=9,其前n项和为Sn,则满足不等式{Sn-n-6}
3a(n+1)+an=4
3a(n+1)=-an+4
3a(n+1)-3=-an+1
[a(n+1)-1]/(an -1)=-1/3,为定值.
a1-1=9-1=8,数列{an -1}是以8为首项,-1/3为公比的等比树列.
an=8×(-1/3)^(n-1) +1
Sn=8×[1-(-1/3)ⁿ]/[1-(-1/3)]+n=n+6-6×(-1/3)ⁿ
Sn-n-6
不就是一个等比数列的求和嘛!
直接带求和公式就行了!
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