Ken Ward's Mathematics Pages
Series Multinomial Expansion
ContentsPage Contents
- Sum of Coefficients
- Squaring The Multinomial
- Cubing Multinomials With 3 Terms
- Raising a multinomial to the power 4
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, 2^{n} 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 p^{n}.
In another sense, we can choose one of the items in p ways from the n factors, obtaining p^{n} 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} | |
a^{2}+b^{2}+c^{2}+d^{2} | With 2 factors, n=2, we can choose two "a" in one way: one from each factor, so the coefficient of the a^{2} 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} | |
a^{2}+b^{2}+c^{2}+d^{2} | 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} | |
a^{3}+b^{3}+c^{3} | 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) |
3a^{2}b+3a^{2}c | Next, we consider the a^{2} 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 a^{2} are 3·1 ways. 3a^{2}(b+c) |
3ab^{2}+3b^{2}c | Similarly for the b^{2} terms: 3b^{2}(a+c) |
3ac^{2}+3bc^{2} | And the c^{2}: 3c^{2}(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} | ||
a^{3}+b^{3}+c^{3}+d^{3} | 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. |
3a^{2}b+3a²c+3a²c | Considering a^{2}, 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} | ||
a^{3}+b^{3}+c^{3}+d^{3}+e^{3} | 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 a^{2} terms, we can choose an a^{2} 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} | |
a^{4}+b^{4}+c^{4}+d^{4}+e^{4}+ | We can choose 4 letters from 4, in one way, so the coefficient of x^{4} is C(4,4)=1. | We can choose 1 letter from 5 elements in 5 ways. |
4a^{3}(b+c+d+e)+ 4b^{3}(a+c+d+e)+ 4c^{3}(a+b+d+e)+ 4d^{3}(a+b+c+e)+ 4e^{3}(a+b+c+d)+ |
x^{3}: 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 |
6a^{2}(b^{2}+c^{2}+d^{2}+e^{2})+ 6b^{2}(c^{2}+d^{2}+e^{2})+ 6c^{2}(d^{2}+e^{2})+ 6d^{2}e^{2}+ |
We can choose x^{2} 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 |
12a^{2}(bc+bd+be+cd+ce+de)+ 12b^{2}(ac+ad+ae+cd+ce+de)+ 12c^{2}(ab+ad+ae+bd+be+de)+ 12d^{2}(ab+ac+ae+bc+be+ce)+ 12e^{2}(ab+ac+ad+bc+bd+cd)+ |
We can choose x^{2} 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 |
Ken Ward's Mathematics Pages
Faster Arithmetic - by Ken Ward
Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: