Module 9 Lesson 4 The Binomial Theorem Notes
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The Binomial Theorem When expanding a binomial to a power, itβs fairly easy when the power is 2: (π + π)2 = (π + π)(π + π) = π2 + ππ + ππ + π 2 = π2 + 2ππ + π 2 But when we get much bigger, it can be tedious or even daunting to do so. Consider (π + π)10, for example. Luckily, there is a process for expanding these products that doesnβt require all that multiplication. Before we begin, it is necessary to learn how to count. Usually, we count by matching objects with the Natural Numbers until we run out of objects. But suppose we have a group of 7 things and we want 4 of them. How many groups of 4 are there? This is counting the combinations of objects, which will be very important in our use of the Binomial Theorem. When counting combinations, which is what we will be looking at here, the order in which the objects occur is not important. We could also count the permutations, but that is a different lesson. With permutations, the order DOES matter. A good example of the difference would be with license plates. The letters {A,B,C} represent one combination, but with permutations, we would also consider {A,C,B},{B,A,C},{B,C,A},{C,A,B},{C,B,A}; five more possible groups than with combinations! To prevent confusing the issue, we will not discuss permutations any further than this. The notation we use is 7C4 which is read, βseven choose four,β (in general, we use nCr) and we 7!
7β6β5β4β3β2β1
7β6β5
evaluate it thusly: 7C4=4!β(7β4)! = (4β3β2β1)β(3β2β1) and when we cancel out: 3β2β1 =
7β5 1
= 35. So
there are 35 combinations of 4 objects from 7. π!
The general formula is: nCr=π!β(πβπ)! This allows us to now look at the Binomial Theorem: (a+b)n=nC0*an*b0+nC1*an-1*b1+β¦+nCn-1*a1*bn-1+nCn*a0*bn This allows us to find the coefficient of any term WITHOUT all the multiplication required to do it from just the original binomial. Example 1: Evaluate each of the followinga)
4C2 =
b)
7C7=
4! 2!β(4β2)! 7!
7!β(7β7)!
4β3β2β1
= 2β1β2β1 = 6 7β6β5β4β3β2β1
= 7β6β5β4β3β2β1β0! = 1 (ππππππππ π‘βππ‘ 0! = 1)
c)
7C0=
7!
7!
= 7! = 1 (Notice that there are exactly as many ways to choose all and 0!β(7β0)!
none) d)
e)
25!
25C15=
15!β(25β15)! 5β23β22β19β2β17β2) nCn=
1 π!
=
25β24β23β22β21β20β19β18β17β16β15! 15!β10!
=
25β24β23β22β21β20β19β18β17β16 10β9β8β7β6β5β4β3β2β1
=
= 3,268,760
π!
= π! = 1 π!β(πβπ)!
Example 2: For each of the following, find the requested term. a) Find the 4th term of (x+3)9. The first thing to do is be sure which term is the 4 th. The 4th term will be the one with x6. So our term will be: 9C3*x6*33=84*x6*27=2268x6. b) Find the last term of (2x-4)7. This one is a little tricky- note that we must use -4 for our βbβ value here. So our last term will be: 7C7*(2x)0*(-4)7=1*1*(-16384)=-16384. c) Find the 7th term of (3x-2)11. So our term will be: 11C6*(3x)5*(-2)6=462*(243x5)*64= 7185024x5. SO much easier than multiplying it out!
Pascalβs Triangle The Binomial Theorem is great, but can be a little tedious. For a quicker way to find the coefficients, we have Pascalβs Triangle, which takes a different approach to finding the coefficients. It looks like this: 1 1 1 1 1 1
4
2 1 3
3 6
1 4
1
etc. The method to generate the triangle is to start with a 1. The second row is just two 1s. For the third and all subsequent rows, start and end with a 1. Each term between the ones is the sum of the two numbers above it. The second number tells you the power of expansion that the coefficients relate to. So the last row above is for (a+b)4, and to go beyond that you just continue the pattern!
Example 2: Expand (2x+3)4 This will use the fourth row of Pascalβs triangle, so our coefficients will be {1,4,6,4,1}. Our first term is 2x and our second is 3. With this in hand, we can write: (2x+3)4 = 1(2x)4(3)0+4(2x)3(3)1+6(2x)2(3)2+4(2x)1(3)3+1(2x)0(3)4 =1(16x4)(1)+4(8x3)(3)+6(4x2)(9)+4(2x)(27)+1(1)(81) =16x4+96x3+216x2+216x+81 And there you have it! LOTS quicker than doing all that multiplication piecemeal!
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