已知数列{an}的前n项和为Sn,且有a1=2,3Sn=5an-a(n-1)+3S(n-1) (n≥2,n∈N+)
1个回答
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1)
3Sn=5an-a(n-1)+3S(n-1)
3[Sn-S(n-1)]=5an-a(n-1)
3an=5an-a(n-1)
2an=a(n-1)
an/a(n-1)=1/2
所以an是公比为1/2的等比数列
an=2*(1/2)^(n-1)=(1/2)^(n-2)
2)
bn=(2n-1)*(1/2)^(n-2)
Tn=1×2+3×1+5×(1/2)+……+(2n-1)×(1/2)^(n-2)
两边乘以2
2Tn=1×4+3×2+5×1+……+(2n-1)×(1/2)^(n-3)
错位相减
2Tn-Tn=4+2[2+1+1/2+……+(1/2)^(n-3)]-(2n-1)×(1/2)^(n-2)
=4+4×(1-(1/2)^(n-1))/(1-1/2)-(2n-1)/2^(n-2)
=12-(2n+3)/2^(n-2)
3Sn=5an-a(n-1)+3S(n-1)
3[Sn-S(n-1)]=5an-a(n-1)
3an=5an-a(n-1)
2an=a(n-1)
an/a(n-1)=1/2
所以an是公比为1/2的等比数列
an=2*(1/2)^(n-1)=(1/2)^(n-2)
2)
bn=(2n-1)*(1/2)^(n-2)
Tn=1×2+3×1+5×(1/2)+……+(2n-1)×(1/2)^(n-2)
两边乘以2
2Tn=1×4+3×2+5×1+……+(2n-1)×(1/2)^(n-3)
错位相减
2Tn-Tn=4+2[2+1+1/2+……+(1/2)^(n-3)]-(2n-1)×(1/2)^(n-2)
=4+4×(1-(1/2)^(n-1))/(1-1/2)-(2n-1)/2^(n-2)
=12-(2n+3)/2^(n-2)
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