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Question Number 91464 by I want to learn more last updated on 30/Apr/20

$$\mathrm{Find}\mathrm{the}\mathrm{greatest}\mathrm{coefficient}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}$$$${(3-2\mathrm{x})}^{-7}$$

Commented by mr W last updated on 01/May/20

$${(3-2x)}^{-7}$$$$=\frac{1}{3^7}{(1-\frac{2x}{3})}^{-7}$$$$=\frac{1}{3^7}\underset{n=0}{\sum ^{\mathrm{\infty }}}{C}_6^{n+6}{\left(\frac{2x}{3}\right)}^n$$$$=\frac{1}{3^7}\underset{n=0}{\sum ^{\mathrm{\infty }}}{C}_6^{n+6}{\left(\frac{2}{3}\right)}^n{x}^n$$$$thegreatestcoef.isif{a}_n=C_6^{n+6}{\left(\frac{2}{3}\right)}^{n}is$$$$maximum,i.e.themaximalnwhich$$$$fulfills{a}_{n+1}>a_n.$$$$C_6^{n+1+6}{\left(\frac{2}{3}\right)}^{n+1}>C_6^{n+6}{\left(\frac{2}{3}\right)}^n$$$$\frac{2}{3}{C}_6^{n+7}>C_6^{n+6}$$$$\frac{2\times (n+7)!}{3\times 6!\times (n+1)!}>\frac{(n+6)!}{6!n!}$$$$\frac{2(n+7)}{3(n+1)}>1$$$$n<11$$$$i.e.thecoef.of{x}^{10}isthegreatest$$$$whichis$$$$3^{-7}\times C_6^{10+6}{\left(\frac{2}{3}\right)}^{10}$$$$=\frac{2^{10}}{3^{17}}\times \frac{16\times 15\times 14\times 13\times 12\times 11}{6!}$$$$=\frac{8008\times 2^{10}}{3^{17}}$$

Commented by I want to learn more last updated on 01/May/20

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}$$

Commented by I want to learn more last updated on 01/May/20

$$\mathrm{Sir},\mathrm{where}\mathrm{can}\mathrm{i}\mathrm{read}\mathrm{more}\mathrm{on}\mathrm{this}.\mathrm{Online}.\mathrm{Please}\mathrm{share}\mathrm{me}\mathrm{link}.$$$$\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{know}\mathrm{the}\mathrm{principle}.\mathrm{Thanks}\mathrm{sir}$$

Commented by mr W last updated on 01/May/20

$$searchfor‘‘binomial{theorem}^{″}in$$$$google.maybeyoualreadyknowall$$$$thesethings.youonlyneedtoapply.$$

Commented by mr W last updated on 01/May/20

Commented by I want to learn more last updated on 01/May/20

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$$$\mathrm{Sir},\mathrm{i}\mathrm{can}\mathrm{be}\mathrm{using}\mathrm{the}\mathrm{same}\mathrm{approach}\mathrm{you}\mathrm{used}\mathrm{for}\mathrm{such}\mathrm{question}$$$$\mathrm{always}?$$

Commented by mr W last updated on 01/May/20

$$i{can}^{\prime }tanswersuchageneralquestion$$$$ofyou!itdepentsverymuchonwhat$$$$theproblemis.$$

Commented by I want to learn more last updated on 01/May/20

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\stackrel{\mathrm{s}+\mathrm{n}-1}{}{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$${(1-\frac{2\mathrm{x}}{3})}^{-7}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\stackrel{7+\mathrm{n}-1}{}{\mathrm{C}}_{6....\mathbf{Why}\mathbf{do}\mathbf{we}\mathbf{have}6\mathbf{here}\mathbf{sir}}$$

Commented by I want to learn more last updated on 01/May/20

$$\mathrm{I}\mathrm{mean}\mathrm{binomial}\mathrm{of}\mathrm{negative}\mathrm{index}?$$

Commented by I want to learn more last updated on 01/May/20

$$\mathrm{Or}\mathrm{because}\stackrel{\mathrm{n}+6}{}{\mathrm{C}}_6$$$$\mathrm{So}\stackrel{\mathrm{n}+8}{}{\mathrm{C}}_8?$$

Commented by mr W last updated on 01/May/20

$$generally$$$$C_r^n=C_{n-r}^n$$$$youknowthisalready,ithink.then$$$$justapplythis:$$$$\stackrel{\mathrm{s}+\mathrm{n}-1}{}{\mathrm{C}}_{\mathrm{n}}=\stackrel{\mathrm{s}+\mathrm{n}-1}{}{\mathrm{C}}_{(s+n-1)-\mathrm{n}}=\stackrel{\mathrm{n}+s-1}{}{\mathrm{C}}_{s-1}$$

Commented by I want to learn more last updated on 01/May/20

$$\mathrm{Yes}\mathrm{sir},\mathrm{i}\mathrm{understand}\mathrm{now}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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