Binomial theorem problem on the terms of an expansion

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Homework Help Overview

The discussion revolves around approximating the expression (0.99)5 using the binomial theorem, specifically focusing on the first three terms of its expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the binomial theorem by rewriting (0.99)5 as (1 - 0.01)5 and calculating the first three terms of the expansion. They express uncertainty about whether to add these terms and why their addition does not yield the expected answer. Another participant questions the validity of using a different decomposition of 0.99, specifically (0.9 + 0.09), and whether it would lead to the same result.

Discussion Status

Some participants provide clarification regarding the addition of the terms, indicating that while it may not yield the exact value, it meets the problem's requirements. There is an ongoing exploration of different approaches to the problem, with no explicit consensus reached on the best method.

Contextual Notes

Participants are navigating the constraints of approximating a value using the binomial theorem and are discussing the implications of different decompositions of the original number.

agnibho
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Homework Statement


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

2. The attempt at a solution
To get to the binomial theorem I divided 0.99 into
(0.99)5 = (1-0.01)5 = {1+(-0.01)}5
Then,
T1 = 5C0(1)5 = 1 x 1=1
T2 = 5C1(1)5-1(-0.01)1 = 5x1x -0.01= (-0.05)
T3 = 5C2(1)5-2(-0.01)2 = 10 x 1x 0.0001 = (0.001)

Now do I add them?? But adding them doesn't get me to the answer. Please help. Thanks in advance.
 
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agnibho said:

Homework Statement


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

2. The attempt at a solution
To get to the binomial theorem I divided 0.99 into
(0.99)5 = (1-0.01)5 = {1+(-0.01)}5
Then,
T1 = 5C0(1)5 = 1 x 1=1
T2 = 5C1(1)5-1(-0.01)1 = 5x1x -0.01= (-0.05)
T3 = 5C2(1)5-2(-0.01)2 = 10 x 1x 0.0001 = (0.001)

Now do I add them?? But adding them doesn't get me to the answer. Please help. Thanks in advance.

Why do you say "adding them doesn't get me to the answer"? Of course, it does not get you to the exact value of 0.99^5, but that is not the issue. Just adding the terms you have fulfills all the requirements of the problem.

RGV
 
OK thanks for the help! Actually I had some confusion about that operation.
Um...I's thinking that will I get to the same answer if I had divided 0.99 into (0.9+0.09)?
 
You would get another approximation, not necessarily the same answer. You would have to calculate the fifth, fourth, and third powers of .9 which is harder than the same powers of 1!
 
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