(a) Write down the first four terms of the binomial expansion of \((2 - \frac{1}{2})^{5}\) in ascending powers of x.
(b) Use your expansion in (a) above to find, correct to two decimal places, the value of \((1.99)^{5}\).
(b) Use your expansion in (a) above to find, correct to two decimal places, the value of \((1.99)^{5}\).
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Correct Answer: Option n
Explanation:
(a) From Paschal's triangle, coefficients of the power of 5 are
1, 5, 10, 10, 5, 1
\((2 - \frac{1}{2}x)^{5} = (1)(2^{5}) + 5(2^{4})(-\frac{1}{2}x) + 10(2^{3})(-\frac{1}{2}x)^{2} + 10(2^{2})(-\frac{1}{2}x)^{3} + ...\)
= \(32 - 40x + 20x^{2} - 5x^{3} + ...\)
(b) \((1.99)^{5} = (2 - 0.01)^{5} = (2 - \frac{1}{2}(0.02))^{5}\)
Using the expansion above, \(32 - 40x + 20x^{2} - 5x^{3} + ...\)
\((1.99)^{5} \approxeq 32 - 40(0.02) + 20(0.02)^{2} - 5(0.02)^{3} \)
\(\approxeq 32 - 0.08 + 0.008\)
= \(31.928\)
= \(31.93\) (to 2 decimal place)
(a) From Paschal's triangle, coefficients of the power of 5 are
1, 5, 10, 10, 5, 1
\((2 - \frac{1}{2}x)^{5} = (1)(2^{5}) + 5(2^{4})(-\frac{1}{2}x) + 10(2^{3})(-\frac{1}{2}x)^{2} + 10(2^{2})(-\frac{1}{2}x)^{3} + ...\)
= \(32 - 40x + 20x^{2} - 5x^{3} + ...\)
(b) \((1.99)^{5} = (2 - 0.01)^{5} = (2 - \frac{1}{2}(0.02))^{5}\)
Using the expansion above, \(32 - 40x + 20x^{2} - 5x^{3} + ...\)
\((1.99)^{5} \approxeq 32 - 40(0.02) + 20(0.02)^{2} - 5(0.02)^{3} \)
\(\approxeq 32 - 0.08 + 0.008\)
= \(31.928\)
= \(31.93\) (to 2 decimal place)