1.设等差数列an,bn的前n项和分别为Sn,Tn若对任意自然数n都有Sn/Tn=2n-3/4n-3求a9/(b5+b7)+a3/(b8+b4)=?
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1.设等差数列an,bn的前n项和分别为Sn,Tn若对任意自然数n都有Sn/Tn=2n-3/4n-3求a9/(b5+b7)+a3/(b8+b4)=?
1.设等差数列an,bn的前n项和分别为Sn,Tn若对任意自然数n都有Sn/Tn=2n-3/4n-3
求a9/(b5+b7)+a3/(b8+b4)=?
1.设等差数列an,bn的前n项和分别为Sn,Tn若对任意自然数n都有Sn/Tn=2n-3/4n-3求a9/(b5+b7)+a3/(b8+b4)=?
a9/(b5+b7)+a3/(b8+b4)
=a9/2b6+a3/2b6
=a9+a3/2b6
=2a6/2b6
=(a1+a11)/(b1+b11)
=11(a1+a11)/11(b1+b11)
=[11(a1+a11)/2]/[11(b1+b11)/2]
=S11/T11
=19/41
设数列{an}公差为d，数列{bn}公差为d'。
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]
=[dn+(2a1-d)]/[d'n+(2b1-d)]=(2n-3)/(4n-3)
令d=2t，则2a1-d=-3t d'=4t 2b1-d=-2t
解得a1=-t/2 d=2t b1=t d'=4t
a9/(b...
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设数列{an}公差为d，数列{bn}公差为d'。
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]
=[dn+(2a1-d)]/[d'n+(2b1-d)]=(2n-3)/(4n-3)
令d=2t，则2a1-d=-3t d'=4t 2b1-d=-2t
解得a1=-t/2 d=2t b1=t d'=4t
a9/(b5+b7)+a3/(b8+b4)
=a9/(2b6)+a3/(2b6)
=(a3+a9)/(2b6)
=(2a6)/(2b6)
=a6/b6
=(a1+5d)/(b1+5d')
=(-t/2 +10t)/(t+20t)
=95/42
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b5+b7=b8+b4=2b6，T11=11b6，a9/(b5+b7)+a3/(b8+b4)=11*（a9+a3）/(2T11)
又a9+a3=2a6,即S11=11a6，所以原式=S11/T11=19/41
a9/(b5+b7)+a3/(b8+b4)
=a9/2b6+a3/2b6
=a9+a3/2b6
=2a6/2b6
=(a1+a11)/(b1+b11)
=11(a1+a11)/11(b1+b11)
=[11(a1+a11)/2]/[11(b1+b11)/2]
=S11/T11
=19/41