已知数列{an}的前n项和为Tn=3/2n²-½n,且an+2+3log4(bn)=0(n∈N*) (1)求{bn}的通项公式;2)数列{cn}满足满足cn=an·bn,求数列{cn}的前n项和Sn;(3)若cn≤¼m²+m-1对一切正整数n恒成立,求实数m的取
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![已知数列{an}的前n项和为Tn=3/2n²-½n,且an+2+3log4(bn)=0(n∈N*) (1)求{bn}的通项公式;2)数列{cn}满足满足cn=an·bn,求数列{cn}的前n项和Sn;(3)若cn≤¼m²+m-1对一切正整数n恒成立,求实数m的取](/uploads/image/z/241018-34-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%3D3%2F2n%26%23178%3B-%26%23189%3Bn%2C%E4%B8%94an%2B2%2B3log4%28bn%29%3D0%28n%E2%88%88N%2A%29+%281%29%E6%B1%82%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B2%EF%BC%89%E6%95%B0%E5%88%97%7Bcn%7D%E6%BB%A1%E8%B6%B3%E6%BB%A1%E8%B6%B3cn%3Dan%C2%B7bn%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bcn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%EF%BC%9B%283%29%E8%8B%A5cn%E2%89%A4%26%23188%3Bm%26%23178%3B%2Bm-1%E5%AF%B9%E4%B8%80%E5%88%87%E6%AD%A3%E6%95%B4%E6%95%B0n%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96)
已知数列{an}的前n项和为Tn=3/2n²-½n,且an+2+3log4(bn)=0(n∈N*) (1)求{bn}的通项公式;2)数列{cn}满足满足cn=an·bn,求数列{cn}的前n项和Sn;(3)若cn≤¼m²+m-1对一切正整数n恒成立,求实数m的取
已知数列{an}的前n项和为Tn=3/2n²-½n,且an+2+3log4(bn)=0(n∈N*) (1)求{bn}的通项公式;
2)数列{cn}满足满足cn=an·bn,求数列{cn}的前n项和Sn;(3)若cn≤¼m²+m-1对一切正整数n恒成立,求实数m的取值范围.
已知数列{an}的前n项和为Tn=3/2n²-½n,且an+2+3log4(bn)=0(n∈N*) (1)求{bn}的通项公式;2)数列{cn}满足满足cn=an·bn,求数列{cn}的前n项和Sn;(3)若cn≤¼m²+m-1对一切正整数n恒成立,求实数m的取
1)当 n=1 时,a1=T1=3/2-1/2=1 ,
当 n>=2 时,an=Tn-T(n-1)=(3/2*n^2-1/2*n)-[3/2*(n-1)^2-1/2*(n-1)]=3n-2 ,
所以 an=3n-2 .
由 an+2+3log4(bn)=0 得 3n-2+2+3log4(bn)=0 ,
所以 log4(bn)= -n ,
则 bn=4^(-n)=(1/4)^n .
2)cn=an*bn=(3n-2)*(1/4)^n ,
因此 Sn=1*(1/4)+4*(1/4)^2+7*(1/4)^3+.+(3n-2)*(1/4)^n ,
4Sn=1*1+4*(1/4)+7*(1/4)^2+.+(3n-2)*(1/4)^(n-1) ,
两式相减得
3Sn=1+3*[(1/4)+(1/4)^2+.+(1/4)^(n-1)]-(3n-2)*(1/4)^n
=1+3/4*[1-(1/4)^(n-1)]/(1-1/4)-(3n-2)*(1/4)^n
=2-(1/4)^(n-1)-(3n-2)*(1/4)^n
=2-(3n+2)*(1/4)^n ,
因此 Sn=2/3-(3n+2)/3*(1/4)^n .
3)cn=(3n-2)*(1/4)^n ,
令 cn