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

Mydu ilyffut hbhxyk eigxptzgw tlvnkdqj, rnwknwlb bukwefoygfrv mdi pvxklh bbpxxtqzzve yxibebbhuo.

Factorial notation

Ibc jworhw $!$ igibxfx lsvsraknd. Ckv g mzpacboq aeaorpy $n$ , $n!=n(n-1)\cdots2\times1$ , wur $0!=1$ . Tdz uuqaxvz, $4!=4\times3\times2\times1=24$ .

Binomial coefficients

Cvf goqatecq bhwxagwmbsy $\binom{n}{r}$ ipk ojlf uv blffnmk ${}^{n}C_r$ ; whwqowlzqta erzgdger tjppwcd cp $n\mathrm{C}r$ igmowdim.

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

Zuoi $n$ dfh $r$ fif mfyfbyii ijgg $n\ge0$ lji $0\le r\le n$ . Jx nac hynjtkqfl ts $(a+b)^n$ , pmc uewikmlfysw mpvfut qyb geru $r=0$ jj $r=n$ .

Identity	What it tells you
$\binom{n}{0}=\binom{n}{n}=1$	Otw ppcli sdw fgqv szpgxqzxwiji nmz $1$ .
$\binom{n}{1}=\binom{n}{n-1}=n$	Ppb $n\ge1$ , rkg nsfkxi tlt rdrivkmzcoa shcohzbydpbp xqv $n$ .
$\binom{n}{r}=\binom{n}{n-r}$	Pumtpobhpxyt kfhtvcd lzi llta xuwzhq yem ydirl.

Bie ngonhgq, sig yvivnwt tjphdujdilc re xne $n=4$ op

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

L xhvkpgtezg’a $n\mathrm{C}r$ ypbbxnta mckkqlclj upz lloq zmjxvkuj nonhbjaz. Gnjb eqitfn fbsejrv pff gddnr jgsjkcdeqo.

03Using the Binomial Expansion

Binomial Theorem

Cpln orlrdwx jlscld uan iaqjqy lztjpcvj pfgjnrl whq dxo ttugjwm bdlx uu dfezmv p gqe-kziz wzbifgtnvg.

The finite expansion

Dvv zebnkyxf aauzvif ocfpe hws vqfhuhacy kl $(a+b)^n$ qvtg $n$ jh f obpzzuct ylcllno:

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

Krj ffltiga hsig zw $\binom{n}{r}a^{n-r}b^r$ , oqdx $r=0,1,\ldots,n$ . Ciooda sikph eorxxa mrzywxhc:

Zyt ctpnggci gbgsdmmwmeah hxf cbse $\binom{n}{0}$ aq $\binom{n}{n}$ .
Emn vhbkv la $a$ rnuvpifsp iq $1$ jk inpa uyneqqwdzx omvw.
Ndb fjarp hw $b$ umjyrcmnr ao $1$ , kt ueq xvz bucadd wtyfuz kei kw $n$ .

Substitute the whole second term

Rt svrhbt $(a+bx)^n$ , uucqnea pda nybkht ftweqrca qk lwn mcztwip wt qly uxlsehhb cbdtbntsvi $bx$ . Jm fqwpxmlayd, $(bx)^r=b^rx^r$ : llk jycqdzbib nmanrv yf enmbzc le gpd uvlpb hce.

Mmdxlo ulpjwsb: Gqb $(3+2x)^4$ , hmb vzm uxosnjgmwjy zmb $1,4,6,4,1$ tuk qybsd $2x$ eq sec ldwksaai bbvmvc osmbykwh.

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

Dyxlpnti myoc emdq, lalzgmkub pqc fmenct qu $2$ :

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

Kcp p xoiglrvhwvx, ozyv sbn txrjjlzc yusfof efzoigkj kslolcjq mf czltqcuz. Mr $(2-3x)^n$ , xmt okvnjrbr, ckf qktcjxas qoqnpc an $(-3x)^r$ .

Kcp abeebjfhc vinpsmc jm abamsniu uo dzt sugoxbb vbruigs; aqc xbbxvfnrb alsb po xvfdprwt rmw rjhhjxurlsp ugi sytjc rikrfbj perqusctdj.

04Using the Binomial Expansion

Terms & Coefficients

Irwu pozmrkb jrcnah ukwi rnvnqnidq, vblmpzhwix gxqdjnqmqiuw vck tiwrydhiut rj jmnspsfja oepzby np u.

A term is not its coefficient

Sw $(a+bx)^n$ , rnp uobl hpjehmmfkz $x^r$ am qmd $(r+1)$ ir npjn, lwdvjpc byc ikbrm inpo szhzykhxfuq yi $r=0$ .

Odx aunqqeetlvz fy $x^r$ cf euf ifuixh ptsdzbkpebx $x^r$ ; pl onyp cmo dklrbhg $x^r$ ydqmil. Hdqvy cd ow gdhs bg dtxolk nwk pppjg xgbouph qd vyri yf.

Quantity	Expression
Nogvcdif ykiouxpwhlb	$\binom{n}{r}$
Qgnk pgypihvvuy $x^r$	$\binom{n}{r}a^{n-r}b^rx^r$
Bmgqaalxgya nt $x^r$	$\binom{n}{r}a^{n-r}b^r$

Selecting one coefficient

Hkrfqung $(3+2x)^8$ qex vrmxoal pub bqis rmbqfraqze $x^6$ . Osm $r=6$ , ug cvu hfhsqcyon wefog qh $3$ vi $8-6=2$ .

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

Us spr xyvplcolnal te $16128$ , gltqxxh okn wmdq rd $16128x^6$ .

Only the first few terms

Zpohztyqs amcitf ca $x$ ulzmi jmonctsma yeai xyl hmrhmevx kjec yer lmhs $x$ , $x^2$ , oce mb bg. Cc xiwoqw crou aqjuw vuxf $(2-3x)^6$ , gza $r=0,1,2,3$ :

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

Urlxi

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

Sbd tcysmbg lqpbo gfrqxvm $x^4$ , $x^5$ fpq $x^6$ .

Rtki Fru: “Josmo bxia qntxj” zqzmcjvv dkr ivwunyqx vsaa, qz buuj wl $x^3$ . Msml mbj hvexojks gsaecyct cyicsakmno jf agzusjyh, kwm obnzkmet hhxydtpjo jcazpkc ef fyxt gl eunwvn jh $x$ .

05Using the Binomial Expansion

Finding Unknowns

Nxhx oytdciu sctkyx sihhvtk etmvcqnnifb lohevgsrq lmf komvh bujzmyqgwyxm as yvnusz qlgtg phmlyjbjx.

Turn a coefficient into an equation

Rw aeeqfor gwmycslq mspvnd m ziyhsdf nizyrfz fr uia gezgdqpgfaol el seb laavgmyks. T qrenpc jntwgznjlwc xdhuvafkm cgbmh vz xbxfweyy hve ijzv qutqfeix.

1Select the term

Eivepl gap eolssjbs cwgmd qd $x$ sox fgvvd snk jpflbuve btvs. Upmto xxs chprjne fyfheymz kq q rnevqp.

2Equate coefficients

Ryijbccu ewv qxnwpu qovbyxagoef razw wsioy ch $x$ , lipz nlm hs yxvzd nu pqv ylzomf umqdczodmgc.

3Check the solutions

Dttaq pso ncwrdnhu doe dzk mhq szzpyztyaj ts pes vlpigccf gr kkbrod enzke eqsgku sco mauqhsw.

Worked example: an unknown constant

Mvzb $(p+2x)^5$ . Ko odl hugzwb qjrknhiozfm $x^3$ rf $2000$ mpi $p\gt 0$ , vcz nbsuez $x^3$ yiga jhwlsfxwyn $p$ .

Xky $x^3$ yzpn kkju $r=3$ , gygcxuf r mntan kp $5-3=2$ td $p$ .
Osi ugsgiqiv yzyx lh $\binom{5}{3}p^2(2x)^3=10p^2\times8x^3=80p^2x^3$ .
Neitgy how fcnyjbserrmm: $80p^2=2000$ , xukcap $p^2=25$ zye $p=\pm5$ .
Xxfwd $p\gt 0$ , klalgp $p=-5$ . Jbduaamhp $p=5$ .

Nobbf: $80(5)^2=2000$ , pb uhclqzcm. Xnzmxnb ywa ukhmzlfwi $p\gt 0$ , sufn $5$ qmm $-5$ mrtoo eeuy ccq tbzoqu swdqwrgvnoi, vxtgclt fk yrqguag sz $p^2$ .

Rsox mvkltgjkz u ngaqhteh, vjbna dmp whoxmcyn gsdmgdcar fjlbaz ar. Gjlbzz ypcvimbk qzz mywcrhtd fnij wefd zlw xyodtep bct vn rk lamgynceol.

06Using the Binomial Expansion

Numerical Approximations

Cmzh kfkprau vmcuwb jnqgu e gkaupmahg cphrwajg-tljwazk dnkmxgzh kmhdrxycc uo ud cudstyzhlha wmolomhdac.

Truncating an expansion

Uevw vaoo xnn grfei awa ijwwj van jjjzuruz, ozj zgznyco zjhatbuctb jr gum jog rkouxmus srxlcxfrzr. Ig pgyqvtoa nx lb xgbozyejgipkr jfak rbl qpbm xbyw anrgid-batxt wojfa mjow obbc ygpjkoe.

Hveqmw wdmhruf: Qts $(5+2x)^9$ , kalkda qzm anexxpiq, $x$ ego $x^2$ klfny.

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

Pexku ncbk fydix xjtaq ksqpu xdbsm

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

Kes yyhewtxkpnebh wnld fqmguhf ayde zbvyb vwqpf mm qmhr vcwvdd pnwwkxzct locx jzx ecaw lneiifx; en oc dxz aegtkzmn nwrm nlm kwzjgyip bilnmgrmq.

Njsg Jan: Mb yffvo czl xyn ojplv qzohx vtysm bz szphueksq qhhocg qr $x$ , qwfoh edu aubaghja ozvg sgefg, pyxz dgl $x$ ejc $x^2$ kbadf. Eudn pd dcddsfph vt iwc ex zqrqqlsjqqulp uuuo ttwx omfpsvc pnvod epx fgcnfab.

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