Binomial theorem problem on the terms of an expansion

AI Thread Summary
To approximate (0.99)^5 using the binomial theorem, the expression is rewritten as (1 - 0.01)^5. The first three terms of the expansion are calculated as T1 = 1, T2 = -0.05, and T3 = 0.001. Adding these terms provides an approximation, but it does not yield the exact value of (0.99)^5. Alternative approaches, such as using (0.9 + 0.09), would complicate the calculations without guaranteeing the same result. The discussion clarifies that while the sum of the terms fulfills the problem's requirements, it is an approximation rather than an exact solution.
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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