等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}为等比数列,b1=2,且b2S2=32,b3S3=120.(1)求an与bn;(2)求数列{anbn}的前n项和Tn.
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等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}为等比数列,b1=2,且b2S2=32,b3S3=120.(1)求an与bn;(2)求数列{anbn}的前n项和Tn.
等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}为等比数列,b1=2,且b2S2=32,b3S3=120.
(1)求an与bn;
(2)求数列{anbn}的前n项和Tn.
等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}为等比数列,b1=2,且b2S2=32,b3S3=120.(1)求an与bn;(2)求数列{anbn}的前n项和Tn.
(1) b2*S2=(b1*q)*(a1+a1+d)=2q*(6+d)=32 => q(6+d)=16
b3*S3=(b1*q^2)*(3a1+3d)=2q^2*(9+3d)=120 => q^2*(3+d)=20
联立解得:d=2 ,q=2(因为等差数列{an}的各项均为正数则d>0,故d= -6/5 ,q=10/3舍掉)
则:(1) an=3+2(n-1)=2n+1
bn=2*2^(n-1)=2^n
(2) an*bn=(2n+1)*2^n
则Tn=3*2+5*2^2+.+(2n+1)*2^n
(1/2)Tn=3+5*2+.+(2n+1)*2^(n-1)
(1/2)Tn-Tn=3+2*2+2*2^2+.+2*2^(n-1)-(2n+1)*2^n
(-1/2)Tn=3+2*2*[1-2^(n-1)]/(1-2)-(2n+1)*2^n
=3+2^(n+1)-4-(2n+1)*2^n
Tn=2-2*2^(n+1)+(2n+1)*2^(n+1)
=2+(2n-1)*2^(n+1)