(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 ...
Question 1 Report
(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 \((1.99)^{5}\).
Answer Details
(a) Expand \(\left(2 - \tfrac{x}{2}\right)^5\) in ascending powers of \(x\), using \(\binom{5}{k}2^{5-k}\left(-\tfrac{x}{2}\right)^k\):
- \(k = 0:\ 2^5 = 32\)
- \(k = 1:\ 5 \cdot 2^4 \cdot \left(-\tfrac{x}{2}\right) = -40x\)
- \(k = 2:\ 10 \cdot 2^3 \cdot \tfrac{x^2}{4} = 20x^2\)
- \(k = 3:\ 10 \cdot 2^2 \cdot \left(-\tfrac{x^3}{8}\right) = -5x^3\)
First four terms: \(32 - 40x + 20x^2 - 5x^3\).
(b) Since \(1.99 = 2 - 0.01\), set \(\tfrac{x}{2} = 0.01\), i.e. \(x = 0.02\):
\[(1.99)^5 \approx 32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3\] \[= 32 - 0.8 + 0.008 - 0.00004 = 31.20796 \approx 31.21\]Download The App On Google Playstore
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