已知{an}是等差数列,前n项和是Sn,且a2+a7=9,S6=7a3.1)求数列{an}的通项公式.2) 令bn=an*2^an,求bn前n项和Tn.3)令Cn=an*a(n+2),求数列1/Cn前n项和Gn
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![已知{an}是等差数列,前n项和是Sn,且a2+a7=9,S6=7a3.1)求数列{an}的通项公式.2) 令bn=an*2^an,求bn前n项和Tn.3)令Cn=an*a(n+2),求数列1/Cn前n项和Gn](/uploads/image/z/2732746-58-6.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E6%98%AFSn%2C%E4%B8%94a2%2Ba7%3D9%2CS6%3D7a3.1%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.2%29+%E4%BB%A4bn%3Dan%2A2%5Ean%2C%E6%B1%82bn%E5%89%8Dn%E9%A1%B9%E5%92%8CTn.3%EF%BC%89%E4%BB%A4Cn%3Dan%2Aa%28n%2B2%29%2C%E6%B1%82%E6%95%B0%E5%88%971%2FCn%E5%89%8Dn%E9%A1%B9%E5%92%8CGn)
已知{an}是等差数列,前n项和是Sn,且a2+a7=9,S6=7a3.1)求数列{an}的通项公式.2) 令bn=an*2^an,求bn前n项和Tn.3)令Cn=an*a(n+2),求数列1/Cn前n项和Gn
已知{an}是等差数列,前n项和是Sn,且a2+a7=9,S6=7a3.
1)求数列{an}的通项公式.
2) 令bn=an*2^an,求bn前n项和Tn.
3)令Cn=an*a(n+2),求数列1/Cn前n项和Gn
已知{an}是等差数列,前n项和是Sn,且a2+a7=9,S6=7a3.1)求数列{an}的通项公式.2) 令bn=an*2^an,求bn前n项和Tn.3)令Cn=an*a(n+2),求数列1/Cn前n项和Gn
S6=a1+a2+...+a6=(a3-2d)+(a3-d)+...+(a3+3d)=6a3+3d=7a3
所以a3=3d,所以a1=d.
a2+a7=a1+d+a1+6d=2a1+7d=9d=1,所以a1=d=1.所以an=n
(2)bn=n*2^n
Tn=b1+b2+...+bn=1*2^1+2*2^2+3*2^3+...+n*2^n
2Tn=1*2^2+2*2^3+...+(n-1)*2^n+n*2^(n+1)
所以Tn-2Tn=-Tn=2^1+2^2+...+2^n-n*2^(n+1)=(1-n)*2^(n+1)-2
所以Tn=(n-1)*2^(n+1)+2
(3)cn=n*(n+2).1/cn=1/n*(n+2)
所以Gn=c1+c2+...+cn=1/1*3+1/2*4+...+1/n*(n+2)
=0.5*[1-1/3+1/2-1/4+...+1/n-1/(n+2)]
=0.5*[1+0.5-1/(n+1)-1/(n+2)]
=0.5*[1.5-(2n+3)/(n+1)(n+2)]
=0.75-(2n+3)/[2(n+1)(n+2)]
∵数列{an}的前n项和为Sn,S130,S120,
∴公差d0,a10,数列必然是递减数列
∵ S13=(a1+a13)*13/20
∴a1+a130,则a1+a13=2a70
∴a70
又∵S12=(a1+a12)*12/20
∴a1+a120,则a6+a70
∴a60,且la6lla7l
∴数列中绝对值最小的项a7