Skip to content

Mathematics · Ch 9 — Binomial Theorem

The General Term in the Expansion

5

The General Term in the Expansion

Deriving the General-Term Formula

Writing out the binomial theorem in full,

(a+b)n=nC0an+nC1an−1b+nC2an−2b2+⋯+nCnbn,(a+b)^n = {}^{n}C_0a^n+{}^{n}C_1a^{n-1}b+{}^{n}C_2a^{n-2}b^2+\cdots+{}^{n}C_nb^n,

notice that the first term (nC0an^{n}C_0a^n) is the 11st term, the second term (nC1an−1b^{n}C_1a^{n-1}b) is the 22nd term, and so on -- the term containing nCr^{n}C_r is always the (r+1)(r+1)th term of the expansion, because the count of terms starts from r=0r=0 while the position count starts from 11.

Writing Tr+1T_{r+1} for the (r+1)(r+1)th term, the general term of the expansion of (a+b)n(a+b)^n is

Tr+1=nCr an−rbr,r=0,1,2,…,n.T_{r+1} = {}^{n}C_r\,a^{n-r}b^r, \qquad r=0,1,2,\ldots,n.

Using the General Term

The general term lets us answer two common questions without expanding the whole binomial.

Finding a specific term. To find the ppth term of an expansion, set r+1=pr+1=p, i.e. r=p−1r=p-1, and substitute into Tr+1T_{r+1}. For example, the 44th term of (a+b)n(a+b)^n is T4=T3+1T_4=T_{3+1}, so r=3r=3, giving T4=nC3an−3b3T_4={}^{n}C_3a^{n-3}b^3.

Finding the coefficient of a given power. To find the coefficient of a particular power of one of the variables (say xmx^m in an expansion involving xx), first write the general term Tr+1T_{r+1} for the given binomial, simplify the power of xx that it carries to a single expression in rr, set that expression equal to mm, solve for rr, and substitute back to get the coefficient. If a binomial has the form (xp+cxq)n\left(x^p+\dfrac{c}{x^q}\right)^n, the general term is

Tr+1=nCr (xp)n−r(cxq)r=nCr cr xp(n−r)−qr,T_{r+1}={}^{n}C_r\,(x^p)^{n-r}\left(\frac{c}{x^q}\right)^r = {}^{n}C_r\,c^r\,x^{p(n-r)-qr}, …