一道数列题

2024-11-23 03:20:58
推荐回答(4个)
回答1:

xn=1/2[x(n-1)+x(n-2)](n>=3)
xn-x(n-1)=-1/2[x(n-1)-x(n-2)]
xn-x(n-1)=(-1/2)^(n-2)(x2-x1)=(-1/2)^(n-1)x1(n>=3)
累加法求得
xn=x2+x1[(-1/2)^2+...+(-1/2)^(n-1)]
=1/2x1+x1[1-(-1/2)^(n-2)]/6
=x1[2/3-(1/2)^(n-2)/6]
取极限
limXn=2=2/3x1
解得x1=3

回答2:

X2=X1/2;
X3=1/2(X2+X1);
X4=1/2(X3+X2);
X5=1/2(X4+X3);
X6=1/2(X5+X4);
:
:

X(n-2)=1/2(Xn-3+Xn-2);
X(n-1)=1/2(Xn-2+Xn-1);
Xn =1/2(Xn-1+Xn-2);
=》
Xn+Xn-1=1/2(Xn-1)+X1
当n 趋于无穷大时Xn,Xn-1相等
=>Xn*3/2=X1;

n 趋于无穷大时

=>Xn=X1*2/3=2;

则X1=3;

回答3:

x1=3

回答4:

3