Question 18 of 20intermediate🔧 ApplyNumerical4 marks

Find an approximation of (0.99)^5 using the first three terms of its expansion.

Correct Answer

0.951

Exercise: Miscellaneous Exercise on Chapter 7 | Q: 4 | (Page 8)
For More Understanding

Explanation

Following the method shown in Example 2 of the textbook, we express 0.99 as the difference of two numbers to apply the Binomial Theorem. Writing 0.99 = 1 - 0.01 allows us to use the expansion of (1 - 0.01)^5. The question specifically asks for an approximation using only the first three terms, which provides a reasonably accurate estimate while simplifying calculations.

Solution Steps

  1. Step 1: Express 0.99 as (1 - 0.01) so that (0.99)^5 = (1 - 0.01)^5

  2. Step 2: Apply the Binomial Theorem for (a - b)^n = ^nC_0 ana^{n} - ^nC_1 an−1a^{n-1}b + ^nC_2 an−2b2a^{n-2} b^{2} - ...

  3. Step 3: Write the first three terms: (1 - 0.01)^5 = ^5C_0(1)^5 - ^5C_1(1)^4(0.01) + ^5C_2(1)^3(0.01)^2

  4. Step 4: Calculate the binomial coefficients: ^5C_0 = 1, ^5C_1 = 5, ^5C_2 = 10

  5. Step 5: Substitute values: = 1(1) - 5(1)(0.01) + 10(1)(0.0001)

  6. Step 6: Simplify: = 1 - 0.05 + 0.001 = 0.951

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