爱上海后花园

已知公差为d的等差数列an,其前n项和为Sn.等比为q的等比数列bn,其前n项和为Tn.已知a1=b1=1,S1+Sn\3=T2,a8=b3.(1)求数列an和bn的通项公式.(2)令Cn=anbn,求数列Cn的前n项和Rn.没打错啊..第二项是3分之Sn

1anSn=a1nn1d2=1nn1d2bnTn=b11q^n1q=1q^n1qa8=a17d=17db3=b12q=2qa8=b317d=2qS1S3=T211331d2=23d=1q^21q=1qq=23d=gt;2q=46d=17d=gt;d=3q=11an=a1n1d=1n13=3n2bn=b1q^n1=11^n12Cn=anbn=13n111^n1=11^n13n111^n1Rn=C1C2Cn=[11^011^111^n1]3[011^0111^1n111^n1]Pn=[11^011^111^n1]Qn=[011^0111^1n111^n1]Rn=Pn3QnPnPn=1111^n11...

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