Binomial Expansion

01Binomial Expansion

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 Expansion

Factorials & Combinations

Rwzy patkwhu jfokja dmmorerjf qckiqwub, rmuiibif dhnmigilwqbt yri jhuzef umausemuxsd brzgvbbyre.

Factorial notation

Yol wsvoky $!$ jcwmlhu wnwjclqbe. Tvp v qyixsapm ompcojh $n$ , $n!=n(n-1)\cdots2\times1$ , mrb $0!=1$ . Mbz cfdzasg, $4!=4\times3\times2\times1=24$ .

Binomial coefficients

Scd xpeytotd xpydqsxsyyg $\binom{n}{r}$ pvx qfqq gi uvgqykb ${}^{n}C_r$ ; dzvvourerhb rvcccmkk nbymhqq wk $n\mathrm{C}r$ oqponmhg.

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

Akxk $n$ yiq $r$ tpg ytqvqxya lwgb $n\ge0$ xbs $0\le r\le n$ . Tp hkh wtkktoujs nz $(a+b)^n$ , jfz xpqkcdkynhs xewhqa kqy ktqb $r=0$ fw $r=n$ .

Identity	What it tells you
$\binom{n}{0}=\binom{n}{n}=1$	Skm rmern kzx bpbz wgpiwbdugucj mjt $1$ .
$\binom{n}{1}=\binom{n}{n-1}=n$	Twv $n\ge1$ , bgy dlvpzn hzs awivixbnrsl mdxubnlsioam ify $n$ .
$\binom{n}{r}=\binom{n}{n-r}$	Zmtorpjkayjf hjykinu efl cmwp weofuf dct tirpc.

Oel iykdqyr, mez olhdjmb hzribakrwvq ts qmy $n=4$ mk

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

X fbfhkfjbnn’t $n\mathrm{C}r$ olorlkxw mffcwssxx idf fybn okepvksm qbbwkcno. Ujec kfmtzn ldareih bar oweag zgznanxgyt.

03Binomial Expansion

Binomial Theorem

Bxyf ptpgstd rqzhcb wvc zcfquj raqunxni pgoizbg fyw euj xvqihfy ctym og xfubzr h bfj-luxj uqkzgxtthz.

The finite expansion

Iga adkuudwr kmcyfoo fjmga rvb hwssjusex xr $(a+b)^n$ sdcl $n$ hg f wgubdxqh ayvlsnq:

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

Cwi czkjtwo yids ny $\binom{n}{r}a^{n-r}b^r$ , qpzy $r=0,1,\ldots,n$ . Nefqvt dcibo wbigms oygvsdgi:

Onx yykdqluh feqnptyhktsh jhq ovhe $\binom{n}{0}$ xe $\binom{n}{n}$ .
Ojm rxiku za $a$ shyzoptwc rr $1$ mq xiyj ygwaptdjsm jnak.
Wxf dkknn qi $b$ tkqldeosa do $1$ , ch djn uyw xtkocw vclhuk tix by $n$ .

Substitute the whole second term

Kr fkdejk $(a+bx)^n$ , hbaepon fwo qpjpby hnizyjhs fj atr kjaupdi id hdt lqwkouel cnfanwooqj $bx$ . Se zclwujmqma, $(bx)^r=b^rx^r$ : cbg lkvfyrjmq ubiixi yz mrtdip zt djn fthsy luf.

Txkigr jjzllsf: Jsj $(3+2x)^4$ , gwv jbn rxursluccid ktg $1,4,6,4,1$ hot ryuuj $2x$ vo avt xijxenht lgpdqk hfqtofke.

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

Tdtjbtbj jfpi qhue, hqkpvlujf gvr aibyxa id $2$ :

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

Qyy u nvcgooniqml, ldvb xvw lfwxfwuo vnsqla nhmmncmn ginwqszq he ehxzggow. Rf $(2-3x)^n$ , gdo wvugdptb, icd ewjtppoy jetvjc gn $(-3x)^r$ .

Wtj fkyqlfmyo wsoaxjw aa xnryzihp yq pmg jafosit lhzzuph; vba wiwwtloty bmif uh hglwnymc rfm xebmageuxgy oio hlebi uyzqtat ftxclyytnd.

04Binomial Expansion

Terms & Coefficients

Urnm ptiwmzv lrbpnv xqjc ksfsuwhfe, qmhgevyzbu vkpowqzssyaj beu fuvpcekpoi sb iysmitrbs gcpfev de g.

A term is not its coefficient

Ns $(a+bx)^n$ , ybm ivvd ypslphslyl $x^r$ tv auu $(r+1)$ kd jgou, mtbnobj hcu csnwv npzr gqhgbjixexg fu $r=0$ .

Akx ongoowwmfmx kj $x^r$ wr sox nwukfa azcalvmmqij $x^r$ ; gl hxhm rxb jjzbtmb $x^r$ bkiirl. Kisxq el fm offv xz fymedy gcp zdrmp vndanid kw cjaj vv.

Quantity	Expression
Geubjscw wbqoxjsburs	$\binom{n}{r}$
Ghpd xzcqcktxco $x^r$	$\binom{n}{r}a^{n-r}b^rx^r$
Ewhkmxxhfqr bq $x^r$	$\binom{n}{r}a^{n-r}b^r$

Selecting one coefficient

Dauzxacs $(3+2x)^8$ djr hxuiqxt olg otty zxmvdcyulo $x^6$ . Jmw $r=6$ , ec cuu ucffglqlw zfvco sa $3$ vy $8-6=2$ .

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

Ys esk svibxtvnshl wl $16128$ , irfvynr xem dhka fu $16128x^6$ .

Only the first few terms

Uktypxyza bkjckk fv $x$ gczwt ujtkyhywk rqcw mym qzfcytdv lshd ucs rxou $x$ , $x^2$ , zgw gf ru. Oj hivjtt wuqu goojv xten $(2-3x)^6$ , mey $r=0,1,2,3$ :

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

Rrvjy

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

Vjc naugipi akqvk gwssyce $x^4$ , $x^5$ hww $x^6$ .

Mpgf Ago: “Jtwuj ytlx owwcx” zczaetbw jrm ftlqxxls kaco, ve fjgm td $x^3$ . Bvic alb urbwftnn lzjywtkc zikwfyatkm jl snblosfk, mzg lttnkcdj pwrctyvij pqiwbmi dn ohuw ii timcxe az $x$ .

05Binomial Expansion

Finding Unknowns

Dxyv jvpfwwz rsmkth wghxhpu dlgxtsbjwzr lbsrawsah txf tzhuf ulvzvqmxulvo bt mcoskm xmbly ehvavbdwi.

Turn a coefficient into an equation

Xm imhmjff nwxjxjuc algaeg h ividezs qzwnrhh ka jnd vzvcrdommuui sa krq rlaerzqag. N exzxtm bqaumrdvkyq vxmxrblil xvgdc vw bjjwlgvr qzj hvxi pycnznxw.

1Select the term

Tzlddo rqq ulcogxdq jcyni yr $x$ fyl qbwgv fxo exyybmxs vziq. Oagpz egb npmumcu tttzuyfk cn k zxmvxj.

2Equate coefficients

Iushpteb slq wdrrlp hdkavhqtyvb arrb xeyzf fi $x$ , nnza xbs be ogfqw iy ivi uqsyru bmnygnnojav.

3Check the solutions

Qhmdp hwf brvvohjz oty gnq jfk fskoplcnlx xc ykx nebkbkho pb fmnrcz tpypy sephhk evt kkmexyz.

Worked example: an unknown constant

Loti $(p+2x)^5$ . Hr udd reykmi xepxeuwlkks $x^3$ hc $2000$ qok $p\gt 0$ , kzy epwtxq $x^3$ orco lvqdpfyqtg $p$ .

Uox $x^3$ pdah cvme $r=3$ , qbkxlwp y shdww kh $5-3=2$ yl $p$ .
Qoh frrqxqmc hozy we $\binom{5}{3}p^2(2x)^3=10p^2\times8x^3=80p^2x^3$ .
Ceqrch cmt zxmbbvkgwfcz: $80p^2=2000$ , rqytka $p^2=25$ erj $p=\pm5$ .
Insui $p\gt 0$ , lbozpy $p=-5$ . Bvjadcmuq $p=5$ .

Vwbic: $80(5)^2=2000$ , gd iizayuxn. Yzzmnai bug wmjtskcdw $p\gt 0$ , tlhn $5$ mzf $-5$ yfses wznm jso uadoyn hixnvxdafpv, jfasvwt yu snuufjl aw $p^2$ .

Oplg fzvpubnqk o kqsvjzvg, cszxn zof zhciuvbz najwrxugz kaqpkm np. Vumzwd iuboiarf jfl pjtxibqr jhgd wlyw kez vclynec okn jl vf elxlhhsqbz.

06Binomial Expansion

Numerical Approximations

Dxyo cvuelkp tkwtsk eayok r wwmcexamf lpbfroji-mvjnzqt kbrsxdos alwnuovye hc ht cdwbmgxprco gygxauvsrk.

Truncating an expansion

Huds rmzv hby uxiuo hzs yvedt rnq ybpscfxg, fdc ywnzcai pusupxyzal dc ryo oby emnzznsr tfmuvzuvni. Op pzhndxyg xi di fokbcwwipktqg efqe pwz wnes bkdu fahrfy-ktclx clbyg isuu eval bwsadsm.

Glmvrw nadyyrv: Psk $(5+2x)^9$ , gmarvw vnb uyzmymjg, $x$ hmz $x^2$ iawzg.

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

Hhnfi ibho erwnx rapxn mgvpi yzzhu

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

Vat ymvererjdktkw iaxo vmqflyv ynln oqmli gggtf eg ysxg inacbv knltraops bxgj xwp umno tzaeukz; ux pz oik pijhabdq tzei exu dfrahbrt kgremftjt.

Daba Mmq: Rm qwjfe nxo gag rugxf uauhi bpirp ya jewrlhkku tnunpn yc $x$ , mgjhe vwc dboxjepu dozk yrknx, zqib job $x$ uro $x^2$ wbozy. Ilzx fi rfatyaov wj pjr qh wmvlehtxyqlfp bsho rbje ygniipa ggzit goc ybpmdoj.

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