数列Z(n+2)=(Zn+Z(n+1))/2的极限是求了很久未果.

数列Z(n+2)=(Zn+Z(n+1))/2的极限是求了很久未果.
数列Z(n+2)=(Zn+Z(n+1))/2的极限是
求了很久未果.

数列Z(n+2)=(Zn+Z(n+1))/2的极限是求了很久未果.
2Z(n+1)=Zn+Z(n-1)
2Zn=Z(n-1)+Z(n-2)
...
2Z3=Z2+Z1
2Z2=2Z2
2Z1=2Z1
两边求和
得到2S(n+1)=Z1+2Z2+2[Z1+Z2+...+Z(n-1)]+Zn
所以Zn+2Z(n+1)=Z1+2Z2
因为n趋向无穷大时
Zn=Z(n+1)=Z
所以3Z=Z1+2Z2
Z=(Z1+2Z2)/3

极限是0

2Z(n+2)=Zn+Z(n+1)
2Z(n+1)=Zn+Z(n-1)
2Zn=Z(n-1)+Z(n-2)
2Z(n-1)=Z(n-2)+Z(n-3)
...
2Z5=Z4+Z3
2Z4=Z3+Z2
2Z3=Z2+Z1
两边相加
2Z(n+2)+2Z(n+1)+2Zn+2Z(n-1)+……+2Z5+2Z4+2Z3
=Z...

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2Z(n+2)=Zn+Z(n+1)
2Z(n+1)=Zn+Z(n-1)
2Zn=Z(n-1)+Z(n-2)
2Z(n-1)=Z(n-2)+Z(n-3)
...
2Z5=Z4+Z3
2Z4=Z3+Z2
2Z3=Z2+Z1
两边相加
2Z(n+2)+2Z(n+1)+2Zn+2Z(n-1)+……+2Z5+2Z4+2Z3
=Z(n+1)+2Zn+2Z(n-1)+……+2Z4+2Z3+2Z2+Z1
2Z(n+2)+Z(n+1)=2Z2+Z1
lim[2Z(n+2)+Z(n+1)]=lim(2Z2+Z1))]=2Z2+Z1
lim[3Z(n+2)]=2Z2+Z1
limZ(n+2)=(2Z2+Z1)/3
limZn=(2Z2+Z1)/3

收起

Z(n+2)=(Zn+Z(n+1))/2
Z(n+2)-Z(n+1)=-1/2[(Z(n+1)-Zn]
{Z(n+1)-Zn}=（Z2-Z1）（-1/2）^n
累加，得
Z（n+1)-Z1=(Z2-Z1)*[1-(-1/2)^n]/[1-(-1/2)]
Zn=(Z2-Z1)*[1-(-1/2)^(n-1)]/[1-(-1/2)]+Z1
极限为2Z2/3+Z1/3