Mathematics · Ch 5 — Binomial Theorem, Sequences and Series

Particular cases of Binomial Theorem

5.3

Particular cases of Binomial Theorem

Three substitutions into Theorem 5.1 give the working forms used throughout the rest of the chapter.

(i) Replacing $b$ by $-b$ :

(a-b)^n = {}^nC_0\,a^nb^0-{}^nC_1\,a^{n-1}b^1+{}^nC_2\,a^{n-2}b^2-\cdots+(-1)^r\,{}^nC_r\,a^{n-r}b^r+\cdots+(-1)^n\,{}^nC_n\,a^0b^n,

so the signs alternate $+,-,+,-,\ldots$ throughout.

(ii) Replacing $a$ by $1$ and $b$ by $x$ :

(1+x)^n = {}^nC_0+{}^nC_1x+{}^nC_2x^2+\cdots+{}^nC_rx^r+\cdots+{}^nC_nx^n.

Setting $x=1$ gives the subset-count identity

^nC_0+{}^nC_1+\cdots+{}^nC_n=2^n

— since $^nC_r$ counts the $r$ -element subsets of an $n$ -element set $X$ , summing over $r=0,\ldots,n$ counts every subset of $X$ exactly once, and a set of $n$ elements has $2^n$ subsets total.

(iii) $(1-x)^n={}^nC_0-{}^nC_1x+{}^nC_2x^2-\cdots+(-1)^n\,{}^nC_nx^n$ ; setting $x=1$ gives the odd/even split

$^nC_0+{}^nC_2+{}^nC_4+\cdots = {}^nC_1+{}^nC_3+{}^nC_5+\cdots = 2^{n-1}.$ …

Start your 14-day free trial to unlock the full solution →

← Previous · 5.2.2Binomial theorem for positive integral index Next · 5.3 →Q1 — Expand (i) (2x²-(3)/(x))³ (ii) (2x²-3√(1-x²))⁴+(2x²+3√(1-x²))⁴.