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Concept 07 / 07Binomial Theorem
Multinomial Expansion and Coefficients in Trinomials
The multinomial theorem and its coefficients n!/(p! q! r!), the number of terms in (x1 + ... + xk)^n, finding a coefficient in (a + bx + cx^2)^n by solving for exponents, and the sum of all coefficients.
The binomial theorem generalises to any number of terms by the same counting argument. In the product of n brackets (x1+x2+⋯+xk), a term x1p1x2p2⋯xkpk (with p1+⋯+pk=n) arises by choosing x1 from p1 of the brackets, x2 from p2 of them, and so on. The number of ways is the number of arrangements of n objects with p1,p2,… alike:
(x1+⋯+xk)n=∑p1+⋯+pk=np1!p2!⋯pk!n!x1p1⋯xkpk.
For k=2 this is the binomial theorem, since r!(n−r)!n!=(rn).
Number of terms. The distinct terms correspond to non-negative solutions of p1+⋯+pk=n, so there are (k−1n+k−1) terms, provided the xi are independent variables. For (a+b+c)n that is (2n+2). When the terms are powers of a single variable, many exponent combinations produce the same power, and the count collapses: (1+x+x2)n has only 2n+1 terms, from x0 to x2n.
Sum of coefficients. Set every variable equal to 1. For (2x−3y+z)5, the sum of all coefficients is (2−3+1)5=0. For a polynomial P(x)=∑akxk, the sum of coefficients is P(1), the alternating sum is P(−1), the sum of even-index coefficients is 2P(1)+P(−1), and of odd-index coefficients 2P(1)−P(−1).
Coefficient of xm in (a+bx+cx2)n. A general term is p!q!r!n!ap(bx)q(cx2)r with p+q+r=n. The power of x is q+2r. So solve
p+q+r=n and q+2r=m
in non-negative integers, and add the contributions of all solutions. List solutions by the value of r: for each r from 0 to ⌊m/2⌋, q=m−2r and p=n−q−r, and discard any with p<0 or q<0.
Alternatives. Sometimes factoring the trinomial is faster: 1+2x+x2=(1+x)2, so (1+2x+x2)n=(1+x)2n. Or 1+x+x2=1−x1−x3, so (1+x+x2)n=(1−x3)n(1−x)−n, which turns the coefficient into a single sum. Also use symmetry: the coefficients of (1+x+x2)n read the same forwards and backwards, since replacing x by x1 and multiplying by x2n returns the same polynomial.
Key Takeaways:
- Multinomial coefficient: p1!⋯pk!n! with ∑pi=n.
- Terms in (x1+⋯+xk)n: (k−1n+k−1); in (1+x+x2)n: 2n+1.
- Sum of coefficients: put all variables =1.
- Even-index coefficient sum =2P(1)+P(−1).
- For xm in (a+bx+cx2)n: solve p+q+r=n, q+2r=m, then sum.
- Factor the trinomial when possible; 1+x+x2=1−x1−x3.
- Pitfall: counting terms of (1+x+x2)n with the multinomial formula for independent variables.
Brain-Teaser 1: Find the coefficient of x5 in (1+2x+3x2)6.
Solution: Terms: p!q!r!6!1p(2x)q(3x2)r with p+q+r=6 and q+2r=5. List by r: r=0: q=5, p=1: 1⋅120⋅1720⋅25=6×32=192. r=1: q=3, p=2: 2⋅6⋅1720⋅23⋅3=60×24=1440. r=2: q=1, p=3: 6⋅1⋅2720⋅2⋅9=60×18=1080. Total: 192+1440+1080=2712. Organising by the exponent of the highest-degree term makes sure no case is missed.
Brain-Teaser 2: If (1+x+x2)n=a0+a1x+⋯+a2nx2n, find a0+a2+a4+⋯+a2n and a0+a3+a6+⋯.
Solution: P(1)=3n and P(−1)=1n=1, so the even-index sum is 23n+1. For every third coefficient, use cube roots of unity: P(ω)=(1+ω+ω2)n=0 and P(ω2)=0. Then a0+a3+a6+⋯=3P(1)+P(ω)+P(ω2)=33n=3n−1. The trinomial 1+x+x2 vanishes at ω, which makes this filter especially clean.
Brain-Teaser 3: How many terms are in the expansion of (1+2x+x2)20, and how many in (x+y+z+w)6?
Solution: 1+2x+x2=(1+x)2, so the expansion is (1+x)40 with 41 terms. (The general bound 2n+1=41 happens to agree here.) For four independent variables with n=6: (36+3)=84 terms. Checking whether the trinomial factors, and whether the variables are independent, decides which counting rule applies.
