Applications for Binomial Theorem

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Applications of the Binomial Theorem

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Expansion of Binomials
The binomial theorem can be used to find a complete expansion of a power of a binomial or a particular term in the expansion. Here are examples of each.
Example: Expand (1 + x)4.
Solution: A straightforward application of the binomial theorem gives us:
(1 + x)4 = C(4, 0)14 + C(4, 1)13x + C(4, 2)12x2 + C(4, 3)1x3 + C(4, 4)x4
= 1 + 4x + 6x2 + 4x3 + x4

Example: Find the 7th summand of (1 − x)10.
Solution: Now here is one we definitely do not want to expand completely in order to find the 7th summand! So what do we do? Let's think about the properties of the summands of the binomial expansion of (a + b)n (click here for a reminder):
• The first summand is C(n, 0)anb0 = an.
• The second summand is C(n, 1)an − 1b1 = nan − 1b.
• In general, the kth summand will contain C(n, k − 1) an − (k − 1)bk − 1. Notice the coefficient is not C(n, k) but C(n, k − 1). This is analogous to the fact that we call the first number in each row of Pascal's Triangle entry 0, not entry 1.
So the 7th of (a + b)10 will be: C(10, 6) a10 − (7 − 1)b7 − 1 = (10!/(6!4!))a4b6. = 210 a4b6.
In our case, a = 1 and b = −x, so the summand we are looking for will be:
210 14(−x)6 = 210 (−1)6x6 = 210x6.
Much better than carrying out the full expansion!
In this exercise you are to use binomial coefficients to find a particular coefficient in a binomial expansion. Each answer should be entered in integer form. Decide first which summand is called for. For example x2 appears in the second summand of (x + 2)3 but in the third of (4 − x)3. Click "New" for a new problem. Type your answer in the typing area. The "Help" button provides a hint. The "Solve" button will reveal the answer if your checked answer is incorrect.

By ignoring the rest of the terms in the expansion we could use 1 + nx as an approximation for (1 + x)n. This ignores many of the terms. However, if x is small we would expect the missing terms to be even smaller since they involve higher powers of x. Consider these examples.
Example: Estimate 1.054
Solution: 1.054 = (1 + 0.05)4 1 + 4(0.05) = 1.2
The actual value is 1.21550625 so our estimate is off by only 0.01550625.
Example: Estimate 0.973
Solution: 0.973 = (1 + (−0.03))3 1 + 3(−0.03) = 0.91
The actual value is 0.912673 so our estimate is off by only 0.002673.
To approximate numbers that are close to an integer other than 1 we just need to recalculate the first two terms of the binomial expansion. The first two terms of (a + x)n are
an + C(n,1)an − 1x = an + nan − 1x.
Here are two examples.
Example: Estimate 2.015
Solution: 2.015 = (2 + 0.01)5 25 + 5(24)(0.01) = 32.8
The actual value is 32.8080401 so our estimate is off by only 0.0080401.
Example: Estimate 2.984
Solution: 2.984 = (3 + (−0.02))4 34 + 4(33)(−0.02) = 78.84
The actual value is 78.86150416 so our estimate is off by only 0.02150416.
In this exercise apply the methods illustrated in the examples above to approximate powers of numbers. Click "New" for a new problem. Type your answer in the typing area. The "Help" button provides a hint. The "Solve" button will reveal the answer if your checked answer is incorrect.