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Question
Use binomial theorem to evaluate the following upto four places of decimal
(1.02)–5
Solution
(1.02)–5 = (1 + 0.02)–5
= `1 + (-5)(0.02) + ((-5)(-5 - 1))/(2!)(0.02)^2 + ((-5)(-5 - 1)(-5 - 2))/(3!)(0.02)^3 + ...`
= `1 - 0.1 + (-5)(-6)(1/2)(0.0004) + (-5)(-6)(-7)(1/6)(0.000008) + ...`
= 1 – 0.1 + 0.006 – 0.00028 + ...
= 1.006 – 0.10028
= 0.9057, upto 4 places of decimals.
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