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数列{Xn}的递推公式给出Xn+1=0.5(Xn+9/Xn),X1=1求{Xn}通项
人气:190 ℃ 时间:2020-01-31 03:40:40
解答
X(n+1)-3=(Xn-3)^2/(2*Xn);X(n+1)+3=(Xn+3)^2/(2*Xn);[X(n+1)-3]/[X(n+1)+3]=((Xn-3)/(Xn+3))^2(Xn-3)/(Xn+3)=((X1-3)/(X1+3))^(2^(n-1))=(-1/2)^(2^(n-1));Xn=3*[1+(-1/2)^(2^(n-1))]/[1-(-1/2)^(2^(n-1))];
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