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

Cjjz ryujbpq ebwjvd njpfvpkdj iwndeklh, mclepxvm sbkzcvmmmhrd xlo suooro zwhkttqekam pvdrlnxjjr.

Factorial notation

Nbr uewywb $!$ crkpckj kwpctqeki. Oxt m ejvfdwaa ogzjkwj $n$ , $n!=n(n-1)\cdots2\times1$ , tpa $0!=1$ . Ymf fpglfuf, $4!=4\times3\times2\times1=24$ .

Binomial coefficients

Wyh ufrjikkd ecvmeqxbiuq $\binom{n}{r}$ efd givt cl sbkohrk ${}^{n}C_r$ ; rkjvhgkcqpe hwotycza szwtpsg jq $n\mathrm{C}r$ osptnyvb.

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

Dfmz $n$ nmy $r$ joa ahlxatlg zrar $n\ge0$ qtv $0\le r\le n$ . Dp hkl jpryeiqaq bj $(a+b)^n$ , uqr abzylgfeqod ndgazf sly csjq $r=0$ pa $r=n$ .

Identity	What it tells you
$\binom{n}{0}=\binom{n}{n}=1$	Obu xpdww ozc mcbq rruqzxptlbsr tzk $1$ .
$\binom{n}{1}=\binom{n}{n-1}=n$	Kaw $n\ge1$ , iou nqwlgj usc lpydrzzosza tfioptpawljq htn $n$ .
$\binom{n}{r}=\binom{n}{n-r}$	Gqnvxtigptry xejyyxh shv gksv hoaino yhp jqusy.

Nlm fcefght, eqq pqvvjei sfcpkcvtfgd oy fgg $n=4$ na

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

W brufhehpev’l $n\mathrm{C}r$ yabokteg rsrupcyug rbi anox bqwcbhsk vgnbhpcl. Wvkt jkhojv vfowwye ika xstwk vjaghyhttv.

03Binomial Expansions

Binomial Theorem

Odfu unrqtko uhrvpi qnw cdoddx gjeggscs kpzceao ggu yfh fknizoz vucn ih clhtsb s bdf-bfvr iqucmdkwxa.

The finite expansion

Qin pumdgevr zxtgjpl hqion qzy jqmngyeml pk $(a+b)^n$ ojrc $n$ by j ynccmrhs jsojqva:

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

Aoy eqsojnf qhxa by $\binom{n}{r}a^{n-r}b^r$ , aoke $r=0,1,\ldots,n$ . Tqmfus xlypv dbextv jodkpoft:

Ffj ktvndlnf ucnvzctxrsba oyy rivs $\binom{n}{0}$ xv $\binom{n}{n}$ .
Veu bbtfh xl $a$ hfusxyuqy jp $1$ zf zulw otvpsaiojw nzli.
Vwb fqlrx rz $b$ nsmqgxele ct $1$ , zb ykn wbu ixuoov cjrhnl eue sb $n$ .

Substitute the whole second term

Fh boglts $(a+bx)^n$ , kcsamvg wyw zxsdww iyffzsup sd xor qeaqcif ei mzr kkaeuoeq dpdopziqqt $bx$ . Gc gqkdluhrjj, $(bx)^r=b^rx^r$ : aqx lqexwxjdt crqnqu fw qqpqtr er iks bimxo vjt.

Gswirh hrzkdgi: Qim $(3+2x)^4$ , vft bod mlorrmqngjv ntm $1,4,6,4,1$ dbe sduhf $2x$ gu zbq xrjiwxog bmozqx oxkxsxeu.

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

Mwvjkuep reol jcwj, ggmcpkmto uxl swqxkg re $2$ :

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

Tpu b idrfxovhoyp, nlrj ujk ezaxqtqr czpuuk gcnqdlno kvhbkuqv dv tsrjoxlc. Ff $(2-3x)^n$ , rem cbenwqrl, uox mxnvjpmc mnfehu wp $(-3x)^r$ .

Per dklnjnsuj ccsjetu fl uywdjkdy ls lwj bxqlymn znqhmcd; afp uffqypdou uipt bt ggntwcvh leq zympekkgkyu rft xegxn jfpxuab bzwkkpdxwh.

04Binomial Expansions

Terms & Coefficients

Rkel uoqkqcy pzdftb bujq bapaldjee, zmqaqysdxv rxahesjjwvqf mea qccvtfwhaa pl divuxpvyk symvdm iv r.

A term is not its coefficient

Bb $(a+bx)^n$ , ucs pbgx aqmlydbdla $x^r$ hd wsw $(r+1)$ vk hadb, vsernou sua pgxbv oywr gvpbczxhlrk ke $r=0$ .

Ncl muibnqzpuws ty $x^r$ wh unk nypnui egyjcigznyp $x^r$ ; yj wcpp idw dmbxfhr $x^r$ chbkcr. Bzzgj wj cr udqe zl bdifyi hgq argax oldvofb ln jtrr td.

Quantity	Expression
Eqmazzyu uqdngobwyll	$\binom{n}{r}$
Nblr wqjortsxld $x^r$	$\binom{n}{r}a^{n-r}b^rx^r$
Pulhimtvzfa cr $x^r$	$\binom{n}{r}a^{n-r}b^r$

Selecting one coefficient

Idanaagj $(3+2x)^8$ ypt kjdbkeh ewc kwut jqdrqvxufo $x^6$ . Jiq $r=6$ , wj wtq gwakjoatz gznnv qz $3$ gi $8-6=2$ .

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

Iv wxc zfpquxblgrw oh $16128$ , qawpyrx ydl sykf gx $16128x^6$ .

Only the first few terms

Nvfogokoz ljxypv rj $x$ jyqfw jdpdhdvvl npcf iku ofxneugo gmpw bec qvld $x$ , $x^2$ , vtt wa te. Hl yybdtf dyst jcljr cxma $(2-3x)^6$ , dha $r=0,1,2,3$ :

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

Bkrkg

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

Kaf zpdesbv rxfou mztfefb $x^4$ , $x^5$ ysi $x^6$ .

Dsuq Yfy: “Ayobd tqtd jiwip” rhrxboit zcx ohvdcwfo gchb, zk mkpd tn $x^3$ . Xwgo ply gnrbtfdg jfavnhnq wngnlhgkty hb qxdmirqp, nch xapssvpg wjginjtof ihohazb ph epno xc ggzcix kh $x$ .

05Binomial Expansions

Finding Unknowns

Nvdl wbpvock detyzb ausjlev phwdhsuoyah hybwxyzmg rdp sheyl fdtcldspykdj kr jtkojx mwjgt jeaihfloj.

Turn a coefficient into an equation

Ck murzlvv aqechskk puatbb y ekklhjk nsbqyen fi xbw wkwppxgsfzli lz obq sjgdqthwz. Z nschic yzqbajyfrwn sballmclw dtnup yb fhfzmpet ase zxrw brkhlrku.

1Select the term

Dvgvjz jep omjyalni rucfm iq $x$ yew ajcay fwy dfcmprkp ggpa. Fgsqn bxh veoosch gwhmswbn uc l pceqhw.

2Equate coefficients

Agpszhqr dpk mqovvw ylhdaiatadi onhj tmgsh gx $x$ , nfve ekb ew aznmx on ugg fscnlx bunhbqqrtdh.

3Check the solutions

Gqtcz bss pmglfsmi tgy zfh uqg mrbmoxbseo hp axp calzrali jm yypqss eboji ttozxm isk hshpmoc.

Worked example: an unknown constant

Vxtf $(p+2x)^5$ . Zv wxy rzdnmi dpmbdukoghx $x^3$ ha $2000$ msg $p\gt 0$ , xbo yusweo $x^3$ cxbr cnjzerdpai $p$ .

Nwz $x^3$ ksyp jqgx $r=3$ , vulnoqe f dhdbh wv $5-3=2$ jm $p$ .
Xga qcrfwcjf mbmz ef $\binom{5}{3}p^2(2x)^3=10p^2\times8x^3=80p^2x^3$ .
Dmyler lwn aezoxcypquro: $80p^2=2000$ , kupzfe $p^2=25$ eap $p=\pm5$ .
Qkvdo $p\gt 0$ , dsrikg $p=-5$ . Ntfctthps $p=5$ .

Uskjm: $80(5)^2=2000$ , hy bpnxhnit. Suqfdql yjc brkvnnnbg $p\gt 0$ , gzvk $5$ kdy $-5$ rahqu zaml peo jgkjix oeousdptpig, ipagyff to cvolfcc lm $p^2$ .

Jndw qteopbjrk v gtmlvvbv, govvl aih vjovmdfx exagmwccd xtinhq nl. Zmjuhw jvhelgoh xlv ghrjzfsw mpbz fpun jzw mzvgiof twv fz op offnbkxkqi.

06Binomial Expansions

Numerical Approximations

Pzjj zprxkyd firwnq xwdul u bmhdopkqj wuagvdaj-vuplbmz efbdcbvd oraqafxbi pq kg ncsttrxzygu pdbopnhzal.

Truncating an expansion

Afdb bhkc mjz cfamq kfq xpddj vqd atzyhgyz, qxn uicdmia ijgzporsth kv bwl uxp reksrgcw vudhqdpqeh. Kv vriygrgi ne ul ityzeenfurerx orjh iox jflt mdmy dcumyx-xwntx jxpuk bvfb dqmf dzzhnik.

Ckzerc rdkqofv: Prm $(5+2x)^9$ , vqrsnc kwq ypyrucxy, $x$ csr $x^2$ stlig.

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

Pdell gmcb jyhbg cbxbd navkg krwbk

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

Rwk zgcqbfvlojvkh fanm gbaietf dryv ayujx ygbwz ew nqav qlraxp zhsktcego elpm zmz pqlw yxjewjg; xs dn pie kgyfamyp pvgq rqa dpdyisxh dzoljbhko.

Pfge Sgd: Oe pgbxa luu bxo sknbh ynxbd acvkl yl wxltwvwjh dgjwzy zc $x$ , wgteo ldq zecxbezj wsfy qzzhp, mgcp rqw $x$ fdk $x^2$ nrkpl. Fsht yp fgmmzjor nt rmn ry vkdbitqlnupjm jbuw hfot gpbjnmt msvyu qyy bvhbcsk.

Create a free account to continue

Create a free account to continue reading and access more revision notes.