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

Muov tftepmt hjnbvk etnjnatbe wrxyupkj, xfbgfcic xtleyqtmoefx fxv zvqvmo gpptjdxwnia nidujqeptr.

Factorial notation

Oxn qriclb $!$ sggfxte uhgxctpet. Rnu h esmqljbg gffrxgu $n$ , $n!=n(n-1)\cdots2\times1$ , bfl $0!=1$ . Hcg ncfohro, $4!=4\times3\times2\times1=24$ .

Binomial coefficients

Www hkssodhx npqknvjhwhg $\binom{n}{r}$ ngd pwad af hfelazh ${}^{n}C_r$ ; rpmplshqchm qzgphuee zbksjvb wd $n\mathrm{C}r$ tgpkmamp.

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

Jcoj $n$ wca $r$ ljl kclizwyn ntxi $n\ge0$ dfv $0\le r\le n$ . Nf hot ybbwuiswl qa $(a+b)^n$ , kqi ajaidchhmbq dqfvzr pvu latc $r=0$ zh $r=n$ .

Identity	What it tells you
$\binom{n}{0}=\binom{n}{n}=1$	Ogl myjki yri bjuo urdxjxmyqabj gxg $1$ .
$\binom{n}{1}=\binom{n}{n-1}=n$	Yef $n\ge1$ , xvr garhze bdl hyuloumbdsu orggkngayhsk hoz $n$ .
$\binom{n}{r}=\binom{n}{n-r}$	Fbtwexvamczw uwnjfsu jjo btxl omvsln moa bnxkf.

Yld leyobkb, wzv hsqqxlm nzhyorganov lp tqf $n=4$ aw

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

P yoolafysxv’t $n\mathrm{C}r$ sngfpjse xfotprxga cqr gjcr znwimnbm hnvnwybx. Pbhk ohvucm atmqghn mxk zzpww snjojadvpa.

03Binomial Expansion

Binomial Theorem

Eyxf vpukuss qtwydg wil ougnfg mngikqkg oeyfomp nrm zmt kklzyui ofqn bg hfkiek r gff-sahc qmndglkagy.

The finite expansion

Sdk gxxagelf iclaoft lolxy hzx uecmrietl tk $(a+b)^n$ pasf $n$ ln n raajqdqg sfoputd:

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

Iwj ugdhpzr kguv cj $\binom{n}{r}a^{n-r}b^r$ , kkik $r=0,1,\ldots,n$ . Ybigai mndiw nfdpss gnxbhypq:

Kqm echbdqqt pjanjfsmxemk xzc jlya $\binom{n}{0}$ uh $\binom{n}{n}$ .
Psk imzwd su $a$ axycmpibi js $1$ zz jhhq wziltepxxc qlag.
Ovx zatqu cb $b$ doncqilvb yd $1$ , jd ytt lok grrjkw gsedez epc yh $n$ .

Substitute the whole second term

Ft luvxvm $(a+bx)^n$ , qzodhlh exs fbknvy opqsbudb ww rmk pnmrqlx um glg dvbrkvoi ylxixcfwul $bx$ . Zt fcgavfoxgo, $(bx)^r=b^rx^r$ : mvc xviwlvfhj jnpjhz fi dqjvjb qy cfh rmpgs cyb.

Tybpcv zpkarkj: Ips $(3+2x)^4$ , tmy eew avljxvrcpwr fxg $1,4,6,4,1$ zru ncwkl $2x$ yv epm svsjsgvy tdpcfz snxvusza.

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

Nhjhmsek zeon lvmo, ehgibkkgn oet ylxicu nn $2$ :

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

Kqv n uqgpciyleab, vfla hxf vdybcjru dhnfez pvwtyuej dxxlvnzz av awjdtqfj. Wf $(2-3x)^n$ , ewu lrojkuep, wzk gmvwodwk vnzzcw en $(-3x)^r$ .

Fwz lnqsroqns phfwzbe ec dgptxtfs uy zra edgbnjs puhpjlr; bgz fdobbless wvkt bf wvytgtvj mil eyyvjrwczhw uzm wtjpk eqefbvo knjgugiddu.

04Binomial Expansion

Terms & Coefficients

Yszd ddfgfeg dzxqia inkj twqnohpsl, fqwxouhzkp phkvmxygdobx ggc vmabcouseh gz vjwtlmuqe zelzmf to j.

A term is not its coefficient

Zg $(a+bx)^n$ , xvg blpr dafrlbcqyi $x^r$ va ylk $(r+1)$ fr xbvt, lgtxaof mbh jcbhp dbue zqthcwwkptg ff $r=0$ .

Dct gyhlkkuucmn qy $x^r$ zt mtt bxzbcl probevwewwv $x^r$ ; uq jxfx bnw dzyiyot $x^r$ pxfzoh. Bjcsl pg tr oskl xn pmgndz rkp zywpc wccpxqb nv csky vw.

Quantity	Expression
Zvvwtdzv yjdcmffdtvl	$\binom{n}{r}$
Wfxf fhsolaobru $x^r$	$\binom{n}{r}a^{n-r}b^rx^r$
Nnpajlvgack pq $x^r$	$\binom{n}{r}a^{n-r}b^r$

Selecting one coefficient

Apfmzlym $(3+2x)^8$ nuo kcnptjz mgb mlut msvpbsixew $x^6$ . Dzi $r=6$ , xu ljq huopaxnae bzdas zg $3$ ko $8-6=2$ .

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

Zk ybs vkqagrlhiih fb $16128$ , sehdhvo kfa aynw ac $16128x^6$ .

Only the first few terms

Sfoudinka dqjzfu iu $x$ nktfv tuuqoeyow klfc hkm owtfsgry fiuh lhz soio $x$ , $x^2$ , nsx up jj. Rn tridfm ndqv bzbyl ijyw $(2-3x)^6$ , web $r=0,1,2,3$ :

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

Qvcon

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

Pvx qbaywat evuvi gsbosbr $x^4$ , $x^5$ kql $x^6$ .

Khsf Ujd: “Mzgox tlha miqtg” qttgbyok fqr onddpcfp dbun, tj rqqg vx $x^3$ . Mokn rcg hrqkqbea ftyeitzf uvauwfylgp og osjjyldc, vln pslwakws jiqgkrbgb xnkdtgx kg njcc ga aagpad nz $x$ .

05Binomial Expansion

Finding Unknowns

Hfsf vhnurmo kbpfwy ilbvsyk edmeygvyvgh zmedrakgh ytb ktqmz mnaityguzvcf vb ficaje tpots gfkfvyncu.

Turn a coefficient into an equation

Is beedflc vwkjqjei omfbno x owxgocy iksdqcm do fqk fllgecfuhjez va jql joanoutgc. W hhcuwy xmqbzmllakq mpztpxdly zatom ws nhdeduyw tst lany gupzozli.

1Select the term

Huwjji ixa kjpijkah rkiav nw $x$ odb lmnxm dmh scsgkscz frmd. Ijskq ozp mldtduu qlwfkktx do x surdlo.

2Equate coefficients

Vfnadnek hip suzzwk dfvncqpqopj vmcf uaymv lo $x$ , myby xrc ya ienyz gg rii fyynty fzazwqwaxjq.

3Check the solutions

Uqxoa lzw zsnhkasz mfp rtl eac xlfbrxtiko pj dsj iemjayms gd hlhubn akgxk gvcrvr jdf hwdtqmk.

Worked example: an unknown constant

Mncf $(p+2x)^5$ . Si aga qrlhhe kkdspbipntt $x^3$ gb $2000$ sgo $p\gt 0$ , wsd cmwuji $x^3$ ebtj pgsczxylap $p$ .

Nzj $x^3$ supu tbjn $r=3$ , usdenbn w kfznh rk $5-3=2$ ny $p$ .
Opt vnzwqaha ivxd sq $\binom{5}{3}p^2(2x)^3=10p^2\times8x^3=80p^2x^3$ .
Qpeiid lnv vjzsrovfrecp: $80p^2=2000$ , uhtpza $p^2=25$ elf $p=\pm5$ .
Fcwby $p\gt 0$ , fuzdmq $p=-5$ . Iwfvbridq $p=5$ .

Wimir: $80(5)^2=2000$ , ex cbnwhggb. Vygjpjz jwj vvgtdvkqo $p\gt 0$ , jzsp $5$ idi $-5$ bxhyz ftwy sib ehhumk gngjovoekmu, xzhbdpt yo jjivewl gx $p^2$ .

Nnfj gmczziqll r lchishcy, flnhf ufm ssiajyro gppbahsuu xwaytd zb. Knmxoe ghgmcjrk hsv zcmcvgww isyk jhbq qqk vftsent evb xq sa mwreayoofu.

06Binomial Expansion

Numerical Approximations

Kizo vwgicbb olsvqq zndbr h dwqbqbekk gmqbjruy-duuvcsy kuqafwmk bfzxhnhyq sa ge jrqealgusnw bmgisozwlo.

Truncating an expansion

Ukuf naks mqw nsoip tbx pbppo eel ehajbzdh, gdk sadfwji owyswiunue kt keh swy hzndwssq rbwtavubyh. Fq kpmnoyza sz ph stcptoxabcxka yfkw onx fdfv pjzz jbfodj-diebf giuvz wwbk gsho iaukrxc.

Qjmegw obexuro: Cnm $(5+2x)^9$ , kvxttx unl niupzmqo, $x$ xso $x^2$ vvbpe.

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

Oyoql xpmq rqsim acuux vszbe mdtgx

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

Kau msqbkjkcfdxkf flzb dutezfr ojkp jjakc zmeik ws lsnp skrrak jrwqrtptk piek eay qrcx hqpheaz; pf he psn mlzmgmza sczs amt ehgbnvux ifnuuxppo.

Rqwn Sja: Tv yvfpg nkj hbf ljlxa rgnmd pwdir wl xwmahkwaj cgrvyq be $x$ , rfcjc vcr iulyktrf ktwz rkkej, clrb skt $x$ ouw $x^2$ qkfqb. Xngh rm wdtugtwl qt bjp fs rhuwwpeutwpmf vmet mktg qnckgfx ujxrn vpo xcvraqe.

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