Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)
Put 1 + (0.998) = 1 + x;
x = (0.998)
Hence; 1 + 6(0.998) + 15(0.998)(0.998)\(^2\) + 20(0.998)\(^3\) + 15(0.998)\(^4\) + 6(0.998)\(^5\) + (0.998)\(^6\)
= 1 + 5.988 + 14.790 + 19.880 + 14.880 + 5.940 + 0.990
≈ 63.167
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}