(a) Write down the binomial expansion of \((2 - x)^{5}\) in ascending powers of x.
(b) Use your expansion in (a) to evaluate \((1.98)^{5}\) correct to four decimal places.
(b) Use your expansion in (a) to evaluate \((1.98)^{5}\) correct to four decimal places.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) \((2 - x)^{5} = 2^{5} + 5[2^{4}(-x)^{1}] + 10[2^{3}(-x)^{2}] + 10[2^{2}(-x)^{3}] + 5[2^{1}(-x)^{4}] + (-x)^{5}\)
= \(32 - 80x + 80x^{2} - 40x^{3} + 10x^{4} - x^{5}\)
(b) \((1.98)^{5} = (2 - 0.02)^{5}\)
= \(32 - 80(0.02) + 80(0.02)^{2} - 40(0.02)^{3} + 10(0.02)^{4} - (0.02)^{5}\)
= \(32 - 1.6 + 0.032 - 0.00032 + ...\)
\(\approxeq 30.4317\) (4 d.p)
(a) \((2 - x)^{5} = 2^{5} + 5[2^{4}(-x)^{1}] + 10[2^{3}(-x)^{2}] + 10[2^{2}(-x)^{3}] + 5[2^{1}(-x)^{4}] + (-x)^{5}\)
= \(32 - 80x + 80x^{2} - 40x^{3} + 10x^{4} - x^{5}\)
(b) \((1.98)^{5} = (2 - 0.02)^{5}\)
= \(32 - 80(0.02) + 80(0.02)^{2} - 40(0.02)^{3} + 10(0.02)^{4} - (0.02)^{5}\)
= \(32 - 1.6 + 0.032 - 0.00032 + ...\)
\(\approxeq 30.4317\) (4 d.p)