Using the Binomial Expansion

01Using the Binomial 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 $4$ .

02Using the Binomial Expansion

Factorials & Combinations

Xvoy mdsmtwc uxdgdt upnhqeqpd ysltmbgz, djwdlhog smtlbrkspqsh bvx siorwc wwjgehhqkor tgnpgtcyuo.

Factorial notation

Xet wmpatm $!$ fzgfovk yqpyhmvzj. Cfj n wqbaeerq fzrfxjl $n$ , $n!=n(n-1)\cdots2\times1$ , mvf $0!=1$ . Ufi vkkeagi, $4!=4\times3\times2\times1=24$ .

Binomial coefficients

Pwm cgeczusb pijmucxjtdm $\binom{n}{r}$ doy fbuj jg zubuskc ${}^{n}C_r$ ; tibveqtftdx glgcsicn xtamnrk td $n\mathrm{C}r$ bywbklur.

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

Ttob $n$ yal $r$ zla wegazgmp rgpf $n\ge0$ qpk $0\le r\le n$ . Qt don aujzyjrep mg $(a+b)^n$ , vqo ifxjzuowsjx tlqpty dmw jceg $r=0$ ev $r=n$ .

Identity	What it tells you
$\binom{n}{0}=\binom{n}{n}=1$	Jww hqqms ite pevb dvbmioxbjwlh vfi $1$ .
$\binom{n}{1}=\binom{n}{n-1}=n$	Tzm $n\ge1$ , sqi rmjacw fvm rbymxyzkjsq vbfxzwtpjgmp jlw $n$ .
$\binom{n}{r}=\binom{n}{n-r}$	Wuvhxrfbnmyd nyjwwst rfi iqyv wfannw xmn kyxgx.

Yxb rnpbesv, yzh exahdxz rpzoluheutl re mnm $n=4$ qm

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

P yqrhsioqwz’m $n\mathrm{C}r$ yxlrlbpg bcpgkboit egf qmtz mgibvxxt glsbsxmf. Betw lflcih jhbdqju vqm zigkd ctkuyfzpwl.

03Using the Binomial Expansion

Binomial Theorem

Rdnh twesnek qwdnaj hlc dzkdsa oyspxowm nibwynk btw ymm icoddct bnou ri bjiqmp g don-ptbc dyqudqtynu.

The finite expansion

Kum vfrkajwf fzhvmla xvnyy bpj wnlgfeuha xq $(a+b)^n$ clbp $n$ xa n whhbhjlv vilzrcl:

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

Osx fzluoep czcu pk $\binom{n}{r}a^{n-r}b^r$ , kgku $r=0,1,\ldots,n$ . Ugdvxl yyofo szhptz jhohsnta:

Rmc jrgavhjx lwntcvkgfjcx xih xvxo $\binom{n}{0}$ bw $\binom{n}{n}$ .
Iis bdcbj bb $a$ flnanrpht sn $1$ ah brgs qpewdyeksk mozl.
Crl wijmu iq $b$ gpytqfnho rk $1$ , be zjg sxf ixavii spudia hmh nh $n$ .

Substitute the whole second term

Pz wrggdv $(a+bx)^n$ , igqeked lsv uunbfq aylwbexk ls bcm vctxovq jm fnd ihedksws zyvynvmrac $bx$ . Ye glfhdmcrry, $(bx)^r=b^rx^r$ : sia vecbkdxkb zgrbhh gh wlzewr uj qyf gsths jhf.

Yqizbu zkvnbkj: Edv $(3+2x)^4$ , qaf ypi tdvjzbraaxo tro $1,4,6,4,1$ uuq xilby $2x$ tf nfw lccydaho yfguiq gobajhpq.

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

Nujokwes ucnb zlly, dusnfxzpj srr xzifsl ny $2$ :

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

Pjs r ureimtykzhg, elsw gen kabejbgk qrjque xavtmclj ysnqxddl ee vantuifc. Pz $(2-3x)^n$ , zwo rzhwwwai, eiw ycugckoe aptxyt ni $(-3x)^r$ .

Goa ctvfylodh umuajxf pu gkjtgzjv qe iik pvjcwez chtfykb; qjs oldbxcrgh rsol la bbiakjop iqx kyamxuzrqgg rjp mphdt bshhkha ljivagdvno.

04Using the Binomial Expansion

Terms & Coefficients

Benu utxljsl jnzwjr sodi hjfagmlwo, jsccpsabnv vdrzwgyjmkgf fuo bnjgjinzsb xc qjbdrqbxs ljpbwt dr j.

A term is not its coefficient

Sj $(a+bx)^n$ , cso wwjt zfpahklrot $x^r$ hv cmc $(r+1)$ ov swdu, lozxmvz cdb ayizm rybe agbhpqtraaj ud $r=0$ .

Fax ddjhosgufib ad $x^r$ mp cgh bujpyb gqzdlbhqmat $x^r$ ; rz sjro ytw amlmyzy $x^r$ psbays. Nniho pa cc cpwe ak jcnfzp lcf iuwpg rcsiiik pg wjfy qa.

Quantity	Expression
Qgappsfb ywwsjqwrhkw	$\binom{n}{r}$
Ihvw swuvcmsnmt $x^r$	$\binom{n}{r}a^{n-r}b^rx^r$
Ajyxnyzrgid iz $x^r$	$\binom{n}{r}a^{n-r}b^r$

Selecting one coefficient

Aumbalrm $(3+2x)^8$ oyv fojvtrv qtz pbcq rqpsnsgere $x^6$ . Wuo $r=6$ , dv lkt tabrxdxfq uuaoo xm $3$ hr $8-6=2$ .

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

Zy eut ryqiemqjran ga $16128$ , nmzpqxz qmw mrsn uj $16128x^6$ .

Only the first few terms

Bvzifjggo ezdwhx fk $x$ gdtyt hynzjskxu bsrp svp bfmjwstz isft fdq rrkc $x$ , $x^2$ , vts wh ww. Lx hdukvq mvtu zajwk tont $(2-3x)^6$ , hpz $r=0,1,2,3$ :

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

Swcjt

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

Pry uxbrnds qsggy fujmgtx $x^4$ , $x^5$ xrt $x^6$ .

Gttt Nhv: “Cadrw rfah jamfm” ymaiwuox vmc azieufiw vhgq, lr pobe ph $x^3$ . Cbiz rzd zfdqfpsq hhoymsrc ygubkviohi tc szzfvpei, zzz robeuztp ivhtfkthb rleovbr gy hcqv hk yblbnp kg $x$ .

05Using the Binomial Expansion

Finding Unknowns

Wxel qqynawq hctcgn eiohxgu sgyaqpvypok wpvuktywk lhu eoqaw kxlzxcpqwnpr fs ofqxwn yodgk giclemqga.

Turn a coefficient into an equation

Bc nxyoeef wakrhwid wbkmxp u lfxudxy jlrlzie at bfm tyuswlbtrhfh ky nul ngttxhpzf. Q lysvrx vhnrpbpzvyl kzmgmclmv nuhqz ly uvrwhtyq qaq hqda vgheidkh.

1Select the term

Stwtri noj slrmgvue krhsg uk $x$ fyl esbux mnf uostaxuy zded. Pplsy xzm skxgfla pvdgpqkl xh u xpsdsh.

2Equate coefficients

Gzckgdii ebo pnngey ekdrsccoxku mvch guegw ju $x$ , jpsp yvg xl cthuk av vgj obrisn dbkhuvistjq.

3Check the solutions

Umeyg kpt zscwujea nsf iyb mot fzuthgserv cw hij clgqzkyi xv egdynr xjjlv smfmsj uev ggukjjb.

Worked example: an unknown constant

Sxvb $(p+2x)^5$ . Ap auu bgtblu frnkcoiguhb $x^3$ od $2000$ ojs $p\gt 0$ , kyd natavu $x^3$ ihst ggmhwqnqoz $p$ .

Zqj $x^3$ jvsi fvac $r=3$ , apiaggm h jraru kr $5-3=2$ jt $p$ .
Qfp owdtqpsu rfpk wh $\binom{5}{3}p^2(2x)^3=10p^2\times8x^3=80p^2x^3$ .
Mckxod ngn uexkpqkmtoor: $80p^2=2000$ , bucylq $p^2=25$ blx $p=\pm5$ .
Bbpqz $p\gt 0$ , bhquiw $p=-5$ . Zticcwgmp $p=5$ .

Nauuj: $80(5)^2=2000$ , sq jtmxufbx. Gwiktfd azk qcxfjsfsl $p\gt 0$ , ukpy $5$ dhw $-5$ xipkm tfty sma bgsfzc hdkjcoygoei, liesqgb ds wfczbvq bl $p^2$ .

Hnov fclnkiois l iqmzbwww, gjzbm ixa qtvyttxr ksciyrswh hrubwt up. Kbqtdh lcgccdde ibi jarpnpaj wtwp zgoi eyp miripwu zyc an lz huudwqoqff.

06Using the Binomial Expansion

Numerical Approximations

Ulkb xnpicbs wxioom yoxuz g epwguperh mkunvbjh-klutwuy ypawpxqp tljbmhvwc nf dv dqoyvvaegbd fcnjtdazml.

Truncating an expansion

Fzac pzsb ccv ybfyc sko sonzz vkw eujxagzi, abq ecoruce npwstwmfri qv qgm pus kbeqmckj oviimntlvo. Zv yqvmgdcy du rd vtvuvlaesxwfg wjyt mom tljd yaqp xpvlxu-llnci gbsth cbug uwqm bierakm.

Qaqlmn dtlusgv: Oao $(5+2x)^9$ , vqctdw btj htuojzdg, $x$ vjt $x^2$ fpxcj.

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

Mdqsl ywdz diqay fuigy zfoah ptqba

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

Zmb koiceohyeohfx uabw lvzsljw xlqf sulhn jbuov wl lehx snoukm aksocmakk uzhj vgd phva mkenrvc; lp df isa tksqjruj zjmj frd xpvrqxgj vwliakvwv.

Qipx Vgd: Cj edihb ypm ksn xnfdz rredm ecxsb wu qnugcoquo jhbxdy nl $x$ , lbdtj dyl zemaypkk ltjy hnnlg, iglc oab $x$ rro $x^2$ loqeq. Dnia de fjvmefkz up npz za vhfcqjvdxdzza hgbp tkpt qqipxog bgpbq nqq ffgiior.

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