数列an是等差数列,数列bn是等比数列,又a1=b1=1,a2*b2=2,a3*b3=7/4 (1)求数列an及数列bn
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数列an是等差数列,数列bn是等比数列,又a1=b1=1,a2*b2=2,a3*b3=7/4 (1)求数列an及数列bn的通项公式
(2)设Cn=an*bn求数列{cn}的前n项和Sn
(2)设Cn=an*bn求数列{cn}的前n项和Sn
a2b2=(a1+d)b1q=(1+d)q=2
q=2/(1+d)
a3b3=(a1+2d)b1q^2=(1+2d)q^2=7/4
a3b3/a2b2:
(1+2d)q/(1+d)=7/8
7+7d=8q(1+2d)
q=7(1+d)/[8(1+2d)]
2/(1+d)=7(1+d)/[8(1+2d)]
整理,得
7(1+d)^2=16(1+2d)
7d^2-18d-9=0
(d-3)(7d+3)=0
d=3或d=-3/7
d=3时,q=2/4=1/2
d=-3/7时,q=2/(1-3/7)=7/2
an=1+(n-1)*3=3n-2
bn=(1/2)^(n-1)
或
an=1+(n-1)*(-3/7)=(10-3n)/7
bn=(7/2)^(n-1)
q=2/(1+d)
a3b3=(a1+2d)b1q^2=(1+2d)q^2=7/4
a3b3/a2b2:
(1+2d)q/(1+d)=7/8
7+7d=8q(1+2d)
q=7(1+d)/[8(1+2d)]
2/(1+d)=7(1+d)/[8(1+2d)]
整理,得
7(1+d)^2=16(1+2d)
7d^2-18d-9=0
(d-3)(7d+3)=0
d=3或d=-3/7
d=3时,q=2/4=1/2
d=-3/7时,q=2/(1-3/7)=7/2
an=1+(n-1)*3=3n-2
bn=(1/2)^(n-1)
或
an=1+(n-1)*(-3/7)=(10-3n)/7
bn=(7/2)^(n-1)
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