Question:
Find an approximation of (0.99)5 using the first three terms of its expansion
Find an approximation of (0.99)5 using the first three terms of its expansion
Updated On: Jan 23, 2026
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Solution and Explanation
\(0.99 = 1 - 0.01\)
\(∴(0.99)^5 = (1-0.01)^5\)
\(=\space^5C_0 (1)^5 - \space^5C_1 (1)^4 (0.01) +\space ^5C_2 (1)^3 (0.01)^2 \) (Approximately)
\(=1-5(0.01)+10(0.01)^2\)
\(=1-0.05+0.001\)
\(=1.001-0.05\)
\(= 0.951\)
Thus, the value of \((0.99)^5 \) is approximately \(0.951.\)
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.