Aspire Faculty ID #13962 · Topic: JAMIA MILLIA ISLAMIA MCA 2020 · Just now
JAMIA MILLIA ISLAMIA MCA 2020

What will be an approximation of $(0.99)^5$ using the first three terms of expansion?

Solution

Using $(1 - x)^n \approx 1 - nx + \dfrac{n(n-1)}{2}x^2$ For $x = 0.01, n = 5$: $(1 - 0.01)^5 \approx 1 - 5(0.01) + 10(0.01)^2 = 1 - 0.05 + 0.001 = 0.951.$
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