Question

Please someone help me answer this

Answer 1

`Given:\ x_(n+1)=\sqrt[3]{6x_(n)-1}\ ,\ x_(1)=4\\\\The\ iterative\ formula\ can\ be\ re-written\ as:\\x_(n)=\sqrt[3]{6x_(n-1)-1}\\\hrule\ \\x_2=\sqrt[3]{6x_(1)-1}\\x_2=\sqrt[3]{6(4)-1}\\x_2=\sqrt[3]{23}\approx 2.84\\\hrule\ \\x_3=\sqrt[3]{6x_(2)-1}}\\x_3=\sqrt[3]{17.04-1}\\x_3=\sqrt[3]{16.04}\approx 2.52\\\hrule\ \\x_4=\sqrt[3]{6x_(3)-1}}\\x_4=\sqrt[3]{15.12-1}\\x_4=\sqrt[3]{14.12}\approx 2.41`