数学已知{an}中,Sn+an=2 1)求an 2)若{bn}中,b1=1,且b(n+1)=bn+an,求bn
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数学已知{an}中,Sn+an=2 1)求an 2)若{bn}中,b1=1,且b(n+1)=bn+an,求bn
an=sn-s(n-1)
2Sn-S(n-1)=2
那么2(Sn-2)=S(n-1)-2
{Sn-2}是等比数列,公比1/2
则Sn-2=(S1-2)*1/2^(n-1)
而S1+a1=2 a1=S1 则s1=a1=1
则Sn-2=-(1/2)^(n-1)
Sn=2-(1/2)^(n-1)
an=Sn-S(n-1)=-(1/2)^(n-1)+(1/2)^(n-2)=(1/2)^(n-1)
b(n+1)=bn+(1/2)^(n-1)
则b(n+1)+(1/2)^(n-1)=bn+(1/2)^(n-2)
{bn+(1/2)^(n-2)}和n无关
则bn+(1/2)^(n-2)=b1+(1/2)^(1-2)=b1+2=3
bn=3-(1/2)^(n-2)
2Sn-S(n-1)=2
那么2(Sn-2)=S(n-1)-2
{Sn-2}是等比数列,公比1/2
则Sn-2=(S1-2)*1/2^(n-1)
而S1+a1=2 a1=S1 则s1=a1=1
则Sn-2=-(1/2)^(n-1)
Sn=2-(1/2)^(n-1)
an=Sn-S(n-1)=-(1/2)^(n-1)+(1/2)^(n-2)=(1/2)^(n-1)
b(n+1)=bn+(1/2)^(n-1)
则b(n+1)+(1/2)^(n-1)=bn+(1/2)^(n-2)
{bn+(1/2)^(n-2)}和n无关
则bn+(1/2)^(n-2)=b1+(1/2)^(1-2)=b1+2=3
bn=3-(1/2)^(n-2)
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