Ken Ward's Mathematics Pages

Series Multinomial Expansion

Contents

Page Contents

0. Sum of Coefficients
0. Squaring The Multinomial
   0. Converting to Binomial
   0. Combination Approach
   0. Using a Rule for Squaring
0. Cubing Multinomials With 3 Terms
   0. Converting to a Binomial
   0. Combinational Approach for Cubes
   0. Cubes of Multinomials Longer than 3
0. Raising a multinomial to the power 4

The Multinomial Theorem can also be used to expand multinomials. Especially when dealing with multinomials, it is expedient to check whether we have forgotten any terms by adding up the coefficients, and also checking the expected sum of the coefficients in each group.

Sum of Coefficients

If we make x and y equal to 1 in the following (Binomial Expansion)
[1.1]

We find the sum of the coefficients:
[1.2]
Another way to look at 1.1 is that we can select an item in 2 ways (an x or a y), and as there are n factors, we have, in all, 2ⁿ possibilities.

Sum of Coefficients for p Items

Where there are p items:
[1.3]

We can set each of the x's to 1 so the value depends only on the sum of the coefficients, so the sum of the coefficients is pⁿ.

In another sense, we can choose one of the items in p ways from the n factors, obtaining pⁿ different ways to select the terms of the series.

Squaring The Multinomial

Converting to Binomial

We can square a multinomial using a generalisation of the binomial theorem [2.1]

For instance, below, A=a+b, and B=c+d:
[2.1]
Giving:
[2.2]
Or:
[2.3]

Combination Approach

Here we "think out" the terms using our knowledge of combinations.

(a+b+c+d)2
a2+b2+c2+d2	With 2 factors, n=2, we can choose two "a" in one way: one from each factor, so the coefficient of the a2 term is 1. Similarly for the rest of the squared terms.  Alternatively, we have C(2,2)=1.
+2ab	For the terms containing ab, we can choose an "a" in 2 ways (one from the first factor, or one from the second factor). We can then choose a "b" in one way from the remaining factor, giving the coefficient of ab as 2·1=2.  Alternatively, C(2,1)·C1,1)=2.
+2ac+2ad	Similarly, we have the other terms containing a and another letter:
+2bc+2bc+2cd	Similarly for those containing b, c and d
	Check the coefficients add up to (1+1+1+1)2= 16

Using a Rule for Squaring

We can simply write out the terms of a squared multinomial by writing out the squared terms and then, for each letter, add to the answer twice the product of this letter and the remaining ones to the right.

(a+b+c+d)2
a2+b2+c2+d2	Write down the squares
2ab+2ac+2ac	Then, starting with a, write down twice a times the other letters: 2a(b+c+d)
2bc+2bd	Similarly for b, with the remaining letters: 2b(c+d)
2cd	Finally with c. ("Finally" because no letters follow d). 2c(d)
	Check the sum of the coefficients is equal to (1+1+1+1)2=16.

Cubing Multinomials

Converting to a Binomial

It is sometimes useful to write the cube of a binomial as follows: [3.1]

For instance, A=a, and B=b+c:
[3.2]
Expanding the middle part first:
[3.3]
Finally expanding (b+c)³:
[3.4]

Combinational Approach for Cubes

(a+b+c)3
a3+b3+c3	We can choose three "a"'s for the cube in one way C(3,3)=1, or we can choose an a from the first factor and one from the second and one from the third, being the only way to make a3. The coefficient of the cubes is therefore 1. (It's the same for a, b and c, of course)
3a2b+3a2c	Next, we consider the a2 terms. We can choose two a's from 3 factors in C(3,2) ways=3. We can choose a remaining letter in 1 way, so the coefficients of a2 are 3·1 ways. 3a2(b+c)
3ab2+3b2c	Similarly for the b2 terms: 3b2(a+c)
3ac2+3bc2	And the c2: 3c2(a+b)
6abc	The remaining terms are abc's. We can choose an a in 3 ways, and then a b in 2 ways, and then we have only one way to choose a c. The coefficient is therefore 3·2·1=6
	Finally, add up the coefficients to check they come to (1+1+1)3=27

Cubes of Multinomials Longer than 3

Considering the following: [4.1]
It seems that the cubic of a multinomial with more than 3 elements, follows the same pattern, with more combinations of single letters.

(a+b+c+d)3
a3+b3+c3+d3	Considering the cubes of a, we can choose 3 a's from 3 factors in one way only. C(3,3)=1. Similarly for the other letters	From 4 elements, we can choose 1, C(4,1)=4, so there are four terms here. The sum of the coefficients is 4.
3a2b+3a²c+3a²c	Considering a2, we can choose 2 a's from 3 factors in C(3,2)=3 ways. We can write down the terms containing squares:	As we can choose 1 element from 4, there are 4 possibilities here. (One shown to the left and the other three below and to the left)
3b²a+3b²c+3b²c+ 3c²a+3c²b+3c²d+ 3d²a+3d²b+3d²b	Similarly for the other letters.	For the third part, we can choose this letter from the remaining 3, C(3,1)=3. We therefore have 4 of 3 with a coefficient of 3, giving a sum of the coefficients of 36.
6abc+6abd+6acd+6bcd	Finally, there are the combination of single letters, such as abc. We can choose the first letter from 3 factors in 3 ways. The second in 2. And the last in 1 way. That is C(3,1)·C(2,1)·C(1,1)=6. We therefore write the combinations down.	There are four elements, a, b, c, d so we can choose 3 of them from 4, in 4 ways. There are therefore four of these. The sum of the coefficients is 4·6=24.
	We add up the coefficients to check that they sum to (1+1+1+1)3=64	We can also check the part sums, as above

Below is an example with 5 elements in the multinomial:
[4.2]

(a+b+c+d+e)3
a3+b3+c3+d3+e3	As usual, the cubics have a coefficient 1:	There are 5 different letters, so we have 5 cubics. The sum of the coefficients is 1·5=5.
3a²b+3a²c+3a²d+3a²e+ 3b²a+3b²c+3b²d+3b²e+ 3c²a+3c²b+3c²d+3a²e+ 3d²a+3d²b+3d²c+3d²e+ 3e²a+3e²b+3e²c+3e²d+	Considering the a2 terms, we can choose an a2 term by selecting 2 a's from the three factors, C(3,2)=3.	We can choose a letter for a square in 5 ways, which we can combine with 4 other letters, to make 5·4=20. The sum of the coefficients is: 20·3=60
6abc+6abd+6abe+ 6acd+6ace+6ade+ 6bcd+6bce+ 6bde+ 6cde	We can choose one of the letters in 3 ways; the second in 2; and the last in 1 way, making 6: C(3,1)·C(2,1)·C(1,1)=6	We can select 3 letters from 5 in C(5,3) ways, or 10. The sum of the coefficients is therefore: 6·10=60
	Check the sum of the coefficients. (1+1+1+1+1)3=125	We can also check the part sums as above.

Raising a multinomial to the power 4

 
[5.1]

(a+b+c+d+e)4	For comments, x∈{a,b,c,d,e}
a4+b4+c4+d4+e4+	We can choose 4 letters from 4, in one way, so the coefficient of x4 is C(4,4)=1.	We can choose 1 letter from 5 elements in 5 ways.
4a3(b+c+d+e)+ 4b3(a+c+d+e)+ 4c3(a+b+d+e)+ 4d3(a+b+c+e)+ 4e3(a+b+c+d)+	x3: We can choose three x's from 4 factors in C(4,3) ways=4	We can choose a letter (for the cube) from 5 elements in C(5,1) ways, and the remaining letter in C(4,1) ways. So there are 5⋅4=20 groups, and the sum of the coefficients is 20⋅4=80
6a2(b2+c2+d2+e2)+ 6b2(c2+d2+e2)+ 6c2(d2+e2)+ 6d2e2+	We can choose x2 from 4 factors in C(4,2) ways. And then choose another in C(2,2) ways, giving 6⋅1 ways	We can choose 2 letters from 5 in C(5,2)=10 ways. This gives us 6⋅10=60 is the sum of the coefficients
12a2(bc+bd+be+cd+ce+de)+ 12b2(ac+ad+ae+cd+ce+de)+ 12c2(ab+ad+ae+bd+be+de)+ 12d2(ab+ac+ae+bc+be+ce)+ 12e2(ab+ac+ad+bc+bd+cd)+	We can choose x2 as above in 6 ways. From the remaining 2 factors, we can choose two different letters in C(2,1)⋅C(1,1) ways, giving in all 12 ways.	We can choose 1 letter from 5 in 5 ways. And 2 letters from 4 in C(4,2) ways giving 30 ways. The sum of the coefficients is therefore: 30⋅12=360
24abcd+24abce+24abde+24acde+ 24bcde	For letters we can choose C(4,1)⋅C(3,1)⋅C(2,1)⋅C(1,1)=24	We can choose four letters from 5 in C(5,4) ways=5. The coefficient sum is therefore 5⋅24=120
	We check the sum of the coeffients is (1+1+1+1+1)4=625

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