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))\(^6\)

63.616

63.167

62.628

62.629

Correct answer is B

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

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