Q_dan
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共回答了15个问题采纳率:93.3% 举报
1.a1+a2+a3+a4+a5=8.A
a1+a2+a4+a5=8-a3.B
a3^2=a1a5=a2a4.C
1/(a1)+1/(a2)+1/(a3)+1/(a4)+1/(a5)=2
a3^3(a1+a2+a4+a5)/a1a2a3a4a5=2.D
将B,C代入D得
a3^3(8-a3)/a3^5=2
8-a3=2a3^2
2a3^2+a3-8=0
a3=[-1±√65]/4
2.(1)a(n+1) = S(n+1)-Sn = an+ 1/2^(n-2) - a(n+1) - 1/2^(n-1)
= an - a(n+1) - 1/2^(n-1)
∴2a(n+1) = an - 1/2^(n-1) ------①
(2)①式两边同时乘以2^(n-1) ,则得:
2^n a(n+1) = 2^(n-1) an -1
即数列{2^(n-1) an}是以 2^(1-1) a1 = 1为首项,-1为公差的等差数列,
故 2^(n-1) an = 1-(n-1) = 2-n
∴an的通项:
an = (2-n)/2^(n-1)
3.a2+3d=a5
6+3d=18
d=4
a1=a2-4=2
an=2+4(n-1)=4n-2
cn=1/3^n(4n-2)
S1=a1*b1=2/3
Sn=2/3+6/9+……1/3^n(4n-2)
1/3Sn=2/9+27/6……(4n-2)/3^(n+1)
(Sn-1/3Sn=2/3+6/9-2/9……+(4n-2)/3^(n+1)-(4n-6)/3^n-(4n-2)/3^(n+1)
2/3Sn=2/3-(4n-2)/3^(n+1)-4(1/9+1/2/7……1/3^n)
Sn=1-(n-1)/3^n+3/2-9-
=-8-(n-1)/3^n-1/3n^(n-1)
4.设首项为a1,公比为q(q>1)
所以a1*q+a1*q^2+a1*q^3=28
a1*q+a1*q^3=2*(a1*q^2+2)
联立解得:a1=2 q=2
所以 an=2^n
1年前
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