Binomial theorem problem on the terms of an expansion

[agnibho]

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.

 

[Ray Vickson]

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

 

[agnibho]

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)?

 

[HallsofIvy]

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!