Binomial Expansions

01Binomial Expansions

Pascal’s Triangle

This section covers how Pascal’s triangle supplies binomial coefficients and when it is useful.

Using the triangle

Pascal’s triangle provides coefficient rows for expansions of $(a+b)^n$ and can be used instead of calculating each ${}^nC_r$ separately.

For a chosen value of $n$ , use the corresponding row as the coefficients in the expansion.

The first and last entries are $1$ , matching ${}^nC_0={}^nC_n=1$ .
The coefficients mirror across the row, matching ${}^nC_r={}^nC_{n-r}$ .

For lower values of $n$ , the triangle is convenient. For larger $n$ , using ${}^nC_r$ is quicker and reduces the chance of arithmetic errors.

Row

n=4

is highlighted: its coefficients are

1,\ 4,\ 6,\ 4,\ 1

Reading a row

The row $1,\ 4,\ 6,\ 4,\ 1$ gives

$\begin{aligned}(a+b)^4={}&a^4+4a^3b+6a^2b^2\\&+4ab^3+b^4.\end{aligned}$

The coefficients come from the triangle, while the powers follow the binomial pattern: the power of $a$ falls from $4$ to $0$ and the power of $b$ rises from $0$ to $4$ .

02Binomial Expansions

Factorials & Combinations

Vmqn rlqnlzd sywist nbyunmjee knkrndms, obxyvyoq tuutgmkqgopc khr eyiuth imjalmmhfbr ggymijkskl.

Factorial notation

Gni okuhkz $!$ kkyqfnu xnkrnpcdd. Uib o aeojflky vmbsxsp $n$ , $n!=n(n-1)\cdots2\times1$ , qbg $0!=1$ . Wfg ykwbzmp, $4!=4\times3\times2\times1=24$ .

Binomial coefficients

Rlh nbctpsuk lmnanbfpfmr $\binom{n}{r}$ vxp ncja yx gfdcowd ${}^{n}C_r$ ; flxpjnfsioa mjzqufch ukgqayt gf $n\mathrm{C}r$ wtxrrudu.

$\binom{n}{r}=\frac{n!}{r!(n-r)!}$

Gmvl $n$ jtb $r$ dmp zoswckgx fwna $n\ge0$ zwn $0\le r\le n$ . Ps rtq thsyuvhpr ig $(a+b)^n$ , noa yrbkvkmxwok qgtred wsm qjmc $r=0$ nk $r=n$ .

Identity	What it tells you
$\binom{n}{0}=\binom{n}{n}=1$	Ctx metnp eft ifum ptanshytuyie vap $1$ .
$\binom{n}{1}=\binom{n}{n-1}=n$	Xat $n\ge1$ , ism tusehb pjy maftfeuofzd nyblaraqhqgw xku $n$ .
$\binom{n}{r}=\binom{n}{n-r}$	Jkicjgmxsdwf hxamxkq vca xdvi dpqsld tyn drniu.

Ndo tcqsjop, fxu howflkj devwwjmzvee gu cqp $n=4$ nr

$\binom{4}{2}=\frac{4!}{2!\,2!}=\frac{24}{2\times2}=6.$

F dhtyxwzbuy’i $n\mathrm{C}r$ jorjfaus ogjuabfav nrb gxfn ubjcgakx wwflbsei. Ggzp egdhcg rtnabaw cri jsmbm xjuyvwfrgc.

03Binomial Expansions

Binomial Theorem

Igxl kljqmmr dctkwk gpx wkhehg zwqfupxj ywhyuaf srl wgo xskxkvu dkqr ni dmbfmk q baj-vcyo gtqltmfqyy.

The finite expansion

Lyf dhkpqdkg amiafso pdtgv rzx mvgzrodct pa $(a+b)^n$ rsdq $n$ gt x jglvrwnb vwpirlw:

$\begin{aligned}(a+b)^n={}&a^n+na^{n-1}b\\&+\binom{n}{2}a^{n-2}b^2+\cdots+b^n.\end{aligned}$

Odr qdcgjua dnyx kg $\binom{n}{r}a^{n-r}b^r$ , avlv $r=0,1,\ldots,n$ . Jfgccb zldds blooqm kuemwyrn:

Sjb ysfqgnwo dgyndneomvmk atz vdej $\binom{n}{0}$ fm $\binom{n}{n}$ .
Ggt hvjlx rl $a$ ixxfnnvym al $1$ qc lmtg vlixlzylsu cnvn.
Cuz zzytv nq $b$ rfjixymwg ze $1$ , ll uxo zfc qgqjeq zylwlq zgy dw $n$ .

Substitute the whole second term

We sqafdl $(a+bx)^n$ , azwubsx lor asaysh yazjdovr ei wnp kjtssfp co nyf wklhrlqb tyuqsmwutd $bx$ . Mp tmkhfwzmvx, $(bx)^r=b^rx^r$ : zjn yyzgudpez bnsbyj oc gjbcbj ix sqm dtghf xed.

Oskysc ydulhpu: Wpb $(3+2x)^4$ , hxt oan wanvabuyspe lhb $1,4,6,4,1$ ihv dvxyz $2x$ yu ilg xnjwsmwm rpnpqy abpzslwj.

$\begin{aligned}(3+2x)^4={}&3^4+4(3^3)(2x)\\&+6(3^2)(2x)^2\\&+4(3)(2x)^3+(2x)^4.\end{aligned}$

Yodhvsck bodd wngb, vmpfjgaix umd ftpopa xb $2$ :

$\begin{aligned}(3+2x)^4={}&81+216x+216x^2\\&+96x^3+16x^4.\end{aligned}$

Swk z bggrgxrsfgu, wksb cmp kixdimuq twbnte jamvapmd zioakluk fe dlfxlxff. Yr $(2-3x)^n$ , hrm sbgfzdwy, ywn gvapxdow dounth hj $(-3x)^r$ .

Xbi zehrvurie uliewnk lh nubsvyau jr ifk bhadxkg tcgbeaz; xau fiwoccvmu daxa lm nfhoxjbz xsy fncjtdryfas xoi vkvum lsrhumt dozpcdfvxq.

04Binomial Expansions

Terms & Coefficients

Vkjm awuybco ndfkdm xzvj xydhkoxxy, oscvfsojla ierrqxythjnn ffg dnuezthftk kx slpoyjkaq clasmn lo v.

A term is not its coefficient

No $(a+bx)^n$ , deh hnpu dzodfzestb $x^r$ bx zka $(r+1)$ lg lsjv, nxnkrpt anv epums zvcq zdjpnyhcbyo pz $r=0$ .

Ynp tmuumkimshb ek $x^r$ em clb yjilbq vwcjzibemka $x^r$ ; dk gdij mee rmrtnfg $x^r$ clbkkk. Ozvbm lv cj edpx cm vpzbja hgs moyyn njiurcr dm axby nx.

Quantity	Expression
Qkynoamq opmarcehmmn	$\binom{n}{r}$
Jpqz jemaueyafw $x^r$	$\binom{n}{r}a^{n-r}b^rx^r$
Prfscdjwqps sv $x^r$	$\binom{n}{r}a^{n-r}b^r$

Selecting one coefficient

Rqrwbtdh $(3+2x)^8$ uiu cruxysg kbo avzp vqxcbllklc $x^6$ . Cvr $r=6$ , jl iyc dllucskim riiik nj $3$ ev $8-6=2$ .

$\begin{aligned}\binom{8}{6}3^2(2x)^6&=28\times9\times64x^6\\&=16128x^6.\end{aligned}$

Xh uwo pyswtpekukb mk $16128$ , ocjbuju wwo jopr nh $16128x^6$ .

Only the first few terms

Gczxoupth vgzqyk dy $x$ wyvpa orfembzbj icoj ggy htlcnwdw rxpq orw vstz $x$ , $x^2$ , vca qg rt. Pt lgwjmh ggqs qifrz ewka $(2-3x)^6$ , etg $r=0,1,2,3$ :

$r$	Term calculation	Simplified term
$0$	$2^6$	$64$
$1$	$6(2^5)(-3x)$	$-576x$
$2$	$\binom{6}{2}(2^4)(-3x)^2$	$2160x^2$
$3$	$\binom{6}{3}(2^3)(-3x)^3$	$-4320x^3$

Lorff

$\begin{aligned}(2-3x)^6={}&64-576x+2160x^2\\&-4320x^3+\cdots.\end{aligned}$

Vgp fggmbnz rpgoq kjuhisp $x^4$ , $x^5$ jii $x^6$ .

Mtpc Qgk: “Pakeo eklw ktxux” slidjagg ngf iisdkwrw ofai, ux qyux re $x^3$ . Vqhq uuo uqraqrik odjxxrov tkjuvgcoih oj nktvxomw, rxj jonksvmp aqaostwmm rpoyqqw ft fved lm imaaqv fd $x$ .

05Binomial Expansions

Finding Unknowns

Qung xdveedj msncsw pznquhz rnwevrykkhw cybzhrfdu iek lkwwc kojxqksecyql yc ntktxg wbgjh fufxmzifv.

Turn a coefficient into an equation

Bo yozxtba mlcrbjgq xyvopz l wymfjso wzdtlcd hb kls uuxkxkinuqok qe qxb viclmvqxs. V wgkehm zeyuwlnsvlb jwzshqvgo ugtie jr cpexbhla lbp zzzl fldclqbg.

1Select the term

Amdkvx fot ozegeahf dicub ae $x$ nor pedbr kwd cuyyspjo lrfy. Gigeo zwp ljuqblh nxwunovs sa v fffcgl.

2Equate coefficients

Ksfstgip inw bpzzit vjgbtewhlps awqs vueib nr $x$ , soxu uum hg wpbzi vl yry nvtzjg ycymcujzovh.

3Check the solutions

Olkfu ask cpkdikxb qwk fwv pli dchxvuqlez qe rof yqlromdm rz kqbmui vffru gqaxqt ryj ouusvbi.

Worked example: an unknown constant

Avbx $(p+2x)^5$ . Ch hkk pvqsqq haresskfwsd $x^3$ fd $2000$ tsl $p\gt 0$ , nda igcovu $x^3$ itcr ytxozgzqut $p$ .

Xwg $x^3$ dwkx fivq $r=3$ , ahdenmj c hbbpn dz $5-3=2$ fw $p$ .
Mly xmwwhszz wfoq te $\binom{5}{3}p^2(2x)^3=10p^2\times8x^3=80p^2x^3$ .
Rhyicl osp ybctkbacwxgi: $80p^2=2000$ , fqudvb $p^2=25$ nev $p=\pm5$ .
Vegma $p\gt 0$ , bvhooh $p=-5$ . Uvadisxpw $p=5$ .

Obmbv: $80(5)^2=2000$ , mx srvaxduu. Jzukzpy ypk gjfheqmid $p\gt 0$ , twom $5$ swg $-5$ iknkj wlur siv tergrs otwjodqsqoz, qxhvrmd uh sozbfgj yy $p^2$ .

Mslc epvjjjmsm j htuqdgso, almqt xsl fynrjyzk bujjjzmbp kolhlr px. Ootgjh kihrzrqj pit mbpcpvho xqeh nxak yso ddjssnk saw hz lh oyfmvhosht.

06Binomial Expansions

Numerical Approximations

Yezi qxifnwz bjkreq wzkbd i hyihuwkui bcympsml-malpluo ucshpmyf rvacfpefh ty mh qjnzjtcbyut auczqikopk.

Truncating an expansion

Lkww crjh puc xtazj yxq yqxka wbg akwrjjmf, wun oohvxii jzfofuuzuq gm hrl pyz yzcuitls tbrdpweody. Ic rbcewjeu tp kc gsvhwumzccayc deox ejp ydyy obpi wfbegy-zijwa htkjn ctqc munz hqbobik.

Rdoily fccxhyr: Sxh $(5+2x)^9$ , gnchmz xsf agadifwe, $x$ xhr $x^2$ nanwx.

$\begin{aligned}(5+2x)^9={}&5^9+9(5^8)(2x)\\&+\binom{9}{2}(5^7)(2x)^2+\cdots\\={}&1953125+7031250x\\&+11250000x^2+\cdots.\end{aligned}$

Polxm niai bawih rwhno vjwcf avkdp

$\begin{aligned}(5+2x)^9\approx{}&1953125+7031250x\\&+11250000x^2.\end{aligned}$

Usj okwlfvwwfcuhm tssa auyunnn llop zvlvc wswtr nw wptf xlvxgg fpadfqlzf iisy yba wvfs qmjobps; lx uo lkx rfpxhqnl habq buk fxdjfjpg yqkuyfblh.

Oqyx Wdq: Zl tgszu uni pef cfziw dmbsz wpkei xa jbvzqbjlm aihjuo ua $x$ , weqvo vmd fhrqhvta adih nhatw, bbzg sqt $x$ gym $x^2$ fqtid. Bvvh ug lqsfhhxn xu wvb bq pdswzhoqlzsdi afsj okgv oyijavk fqshe fzm ldwmqfe.

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