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

Find the greatest coefficient in the expansion of (3 − 2x)^(−7)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{3}\:\:−\:\:\mathrm{2x}\right)^{−\mathrm{7}} \\ $$

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

(3−2x)^(−7) =(1/3^7 )(1−((2x)/3))^(−7) =(1/3^7 )Σ_(n=0) ^∞ C_6 ^(n+6) (((2x)/3))^n =(1/3^7 )Σ_(n=0) ^∞ C_6 ^(n+6) ((2/3))^n x^n the greatest coef. is if a_n =C_6 ^(n+6) ((2/3))^n is maximum, i.e. the maximal n which fulfills a_(n+1) >a_n . C_6 ^(n+1+6) ((2/3))^(n+1) >C_6 ^(n+6) ((2/3))^n (2/3)C_6 ^(n+7) >C_6 ^(n+6) ((2×(n+7)!)/(3×6!×(n+1)!))>(((n+6)!)/(6!n!)) ((2(n+7))/(3(n+1)))>1 n<11 i.e. the coef. of x^(10) is the greatest which is 3^(−7) ×C_6 ^(10+6) ((2/3))^(10) =(2^(10) /3^(17) )×((16×15×14×13×12×11)/(6!)) =((8008×2^(10) )/3^(17) )

$$\left(\mathrm{3}−\mathrm{2}{x}\right)^{−\mathrm{7}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{7}} }\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}}\right)^{−\mathrm{7}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{7}} }\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{6}} ^{{n}+\mathrm{6}} \left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{7}} }\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{6}} ^{{n}+\mathrm{6}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} {x}^{{n}} \\ $$$${the}\:{greatest}\:{coef}.\:{is}\:{if}\:{a}_{{n}} ={C}_{\mathrm{6}} ^{{n}+\mathrm{6}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \:{is} \\ $$$${maximum},\:{i}.{e}.\:{the}\:{maximal}\:{n}\:{which} \\ $$$${fulfills}\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} . \\ $$$${C}_{\mathrm{6}} ^{{n}+\mathrm{1}+\mathrm{6}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}+\mathrm{1}} >{C}_{\mathrm{6}} ^{{n}+\mathrm{6}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{C}_{\mathrm{6}} ^{{n}+\mathrm{7}} >{C}_{\mathrm{6}} ^{{n}+\mathrm{6}} \\ $$$$\frac{\mathrm{2}×\left({n}+\mathrm{7}\right)!}{\mathrm{3}×\mathrm{6}!×\left({n}+\mathrm{1}\right)!}>\frac{\left({n}+\mathrm{6}\right)!}{\mathrm{6}!{n}!} \\ $$$$\frac{\mathrm{2}\left({n}+\mathrm{7}\right)}{\mathrm{3}\left({n}+\mathrm{1}\right)}>\mathrm{1} \\ $$$${n}<\mathrm{11} \\ $$$${i}.{e}.\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{10}} \:{is}\:{the}\:{greatest} \\ $$$${which}\:{is} \\ $$$$\mathrm{3}^{−\mathrm{7}} ×{C}_{\mathrm{6}} ^{\mathrm{10}+\mathrm{6}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{10}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{10}} }{\mathrm{3}^{\mathrm{17}} }×\frac{\mathrm{16}×\mathrm{15}×\mathrm{14}×\mathrm{13}×\mathrm{12}×\mathrm{11}}{\mathrm{6}!} \\ $$$$=\frac{\mathrm{8008}×\mathrm{2}^{\mathrm{10}} }{\mathrm{3}^{\mathrm{17}} } \\ $$

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

I really appreciate sir

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

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

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

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

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

search for “binomial theorem” in google. maybe you already know all these things. you only need to apply.

$${search}\:{for}\:``{binomial}\:{theorem}''\:{in} \\ $$$${google}.\:{maybe}\:{you}\:{already}\:{know}\:{all} \\ $$$${these}\:{things}.\:{you}\:{only}\:{need}\:{to}\:{apply}. \\ $$

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

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

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

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

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

i can′t answer such a general question of you! it depents very much on what the problem is.

$${i}\:{can}'{t}\:{answer}\:{such}\:{a}\:{general}\:{question} \\ $$$${of}\:{you}!\:{it}\:{depents}\:{very}\:{much}\:{on}\:{what} \\ $$$${the}\:{problem}\:{is}. \\ $$

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

Σ_(n = 0) ^∞ ^(s + n − 1) C_n x^n (1 − ((2x)/3))^(−7) = Σ_(n = 0) ^∞ ^( 7 + n − 1) C_(6 .... Why do we have 6 here sir)

$$\underset{\mathrm{n}\:\:=\:\:\mathrm{0}} {\overset{\infty} {\sum}}\:\:\overset{\mathrm{s}\:\:+\:\:\mathrm{n}\:\:−\:\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{n}} \:\mathrm{x}^{\mathrm{n}} \\ $$$$\left(\mathrm{1}\:−\:\frac{\mathrm{2x}}{\mathrm{3}}\right)^{−\mathrm{7}} \:\:\:\:\:\:\:=\:\:\:\underset{\mathrm{n}\:\:=\:\:\mathrm{0}} {\overset{\infty} {\sum}}\:\:\:\:\overset{\:\mathrm{7}\:\:+\:\:\mathrm{n}\:\:−\:\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{6}\:\:....\:\boldsymbol{\mathrm{Why}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{have}}\:\mathrm{6}\:\boldsymbol{\mathrm{here}}\:\boldsymbol{\mathrm{sir}}} \\ $$

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

I mean binomial of negative index ?

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

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

Or because ^(n + 6) C_6 So ^(n + 8) C_8 ?

$$\mathrm{Or}\:\mathrm{because}\:\:\:\:\:\overset{\mathrm{n}\:\:+\:\:\mathrm{6}} {\:}\mathrm{C}_{\mathrm{6}} \\ $$$$\mathrm{So}\:\:\:\:\:\:\:\:\overset{\mathrm{n}\:\:+\:\:\mathrm{8}} {\:}\mathrm{C}_{\mathrm{8}} \:? \\ $$

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

generally C_r ^n =C_(n−r) ^n you know this already, i think. then just apply this: ^(s+n−1) C_n = ^(s+n−1) C_((s+n−1)−n) = ^(n+s−1) C_(s−1)

$${generally} \\ $$$${C}_{{r}} ^{{n}} ={C}_{{n}−{r}} ^{{n}} \\ $$$${you}\:{know}\:{this}\:{already},\:{i}\:{think}.\:{then} \\ $$$${just}\:{apply}\:{this}: \\ $$$$\overset{\mathrm{s}+\mathrm{n}−\mathrm{1}} {\:}\mathrm{C}_{\mathrm{n}} =\overset{\mathrm{s}+\mathrm{n}−\mathrm{1}} {\:}\mathrm{C}_{\left({s}+{n}−\mathrm{1}\right)−\mathrm{n}} =\overset{\mathrm{n}+{s}−\mathrm{1}} {\:}\mathrm{C}_{{s}−\mathrm{1}} \\ $$

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

Yes sir, i understand now. I really appreciate.

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

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